Algebra 3 Quiz: Solving Trigonometric Equations
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Solving Trigonometric EquationsQuestion 1 of 12

Solve tan⁡2x−3=0\tan^2 x - 3 = 0 over the interval [0,π)[0, \pi).

π3\frac{\pi}{3}
π3,2π3\frac{\pi}{3}, \frac{2\pi}{3}
π3,4π3\frac{\pi}{3}, \frac{4\pi}{3}
π3,2π3,4π3,5π3\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}
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Algebra 3 Quiz

Algebra 3 Quiz: Solving Trigonometric Equations

Practice Solving Trigonometric Equations in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Solving Trigonometric Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Solve tan⁡2x−3=0\tan^2 x - 3 = 0 over the interval [0,π)[0, \pi).

  1. π3\frac{\pi}{3}
  2. π3,2π3\frac{\pi}{3}, \frac{2\pi}{3} (correct answer)
  3. π3,4π3\frac{\pi}{3}, \frac{4\pi}{3}
  4. π3,2π3,4π3,5π3\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}
Explanation: When you see a squared trigonometric function set equal to a constant, your first move is to isolate the trig function: tan⁡2x=3\tan^2 x = 3, so tan⁡x=±3\tan x = \pm\sqrt{3}. Now ask two questions: where does the tangent equal 3\sqrt{3}, and where does it equal −3-\sqrt{3}, within the given interval [0,π)[0,\pi)? On the unit circle, tan⁡x=3\tan x = \sqrt{3} at x=π3x=\frac{\pi}{3}. Since tangent is negative in the second quadrant, tan⁡x=−3\tan x = -\sqrt{3} at x=2π3x=\frac{2\pi}{3}. Both of these are inside [0,π)[0,\pi), so the solution set is π3,2π3\frac{\pi}{3}, \frac{2\pi}{3}. Now look at the wrong choices. π3\frac{\pi}{3} alone forgets that taking a square root introduces both positive and negative values, so it misses the negative-tangent solution. π3,4π3\frac{\pi}{3}, \frac{4\pi}{3} incorrectly includes 4π3\frac{4\pi}{3}, which is outside [0,π)[0,\pi), and it also skips 2π3\frac{2\pi}{3}. The four-angle set π3,2π3,4π3,5π3\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3} would be correct over [0,2π)[0,2\pi), but the interval here ends at π\pi, so the last two angles are not allowed. A strong habit for trig equations is to solve algebraically first, then check every candidate against the interval. Drawing a quick unit circle or tangent graph helps you see which quadrants are included — and prevents you from either dropping solutions or carrying in angles outside the requested range.

Question 2

Solve cos⁡(πx)=22\cos(\pi x) = \frac{\sqrt{2}}{2} over the interval 0≤x≤20 \le x \le 2.

  1. 14\frac{1}{4}
  2. 14,74,94,154\frac{1}{4}, \frac{7}{4}, \frac{9}{4}, \frac{15}{4}
  3. 14,34,54,74\frac{1}{4}, \frac{3}{4}, \frac{5}{4}, \frac{7}{4}
  4. 14,74\frac{1}{4}, \frac{7}{4} (correct answer)
Explanation: When you solve cos⁡(πx)=22\cos(\pi x)=\frac{\sqrt2}{2}, you're really solving the general cosine equation cos⁡θ=22\cos \theta=\frac{\sqrt2}{2} with θ=πx\theta=\pi x. Cosine is positive in Quadrants I and IV, so the reference angle π4\frac{\pi}{4} gives θ=π4\theta=\frac{\pi}{4} and θ=7π4\theta=\frac{7\pi}{4} after adding multiples of 2π2\pi. Dividing by π\pi, this becomes x=14+2kx=\frac14+2k or x=74+2kx=\frac74+2k. On 0≤x≤20\le x\le 2, only x=14x=\frac14 and x=74x=\frac74 fit, so the correct answer is 14,74\frac14,\frac74. The single answer 14\frac14 misses the second valid solution in Quadrant IV. The answer 14,74,94,154\frac14,\frac74,\frac94,\frac{15}{4} correctly identifies the pattern but includes 94\frac94 and 154\frac{15}{4}, which are outside the interval 00 to 22. The answer 14,34,54,74\frac14,\frac34,\frac54,\frac74 confuses angles where cosine equals ±22\pm\frac{\sqrt2}{2}: 3π4\frac{3\pi}{4} and 5π4\frac{5\pi}{4} are Quadrant II and III angles where cosine is negative, so they do not satisfy the original equation. A useful habit: when solving trig equations, first find the general solutions using the correct quadrants, then filter them through the given interval. Also, always check whether a listed angle would make the original function positive or negative — many wrong choices come from listing all special angles rather than only the ones in the correct sign regions.

Question 3

Solve tan⁡(2x)=1\tan(2x) = 1 over the interval [0,π)[0, \pi).

  1. π8\frac{\pi}{8}
  2. π4,5π4\frac{\pi}{4}, \frac{5\pi}{4}
  3. π8,5π8\frac{\pi}{8}, \frac{5\pi}{8} (correct answer)
  4. π8,5π8,9π8,13π8\frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8}
Explanation: Whenever you see a trigonometric equation like tan⁡(2x)=1\tan(2x)=1, remember that the tangent function has period π\pi, not 2π2\pi. So tan⁡(θ)=1\tan(\theta)=1 when θ=π4+kπ\theta=\frac{\pi}{4}+k\pi for any integer kk. Here θ=2x\theta=2x, so set 2x=π4+kπ⇒x=π8+kπ2.2x=\frac{\pi}{4}+k\pi \quad\Rightarrow\quad x=\frac{\pi}{8}+\frac{k\pi}{2}. Now find which of these values lie in [0,π)[0,\pi). For k=0k=0, you get π8\frac{\pi}{8}. For k=1k=1, you get π8+π2=5π8\frac{\pi}{8}+\frac{\pi}{2}=\frac{5\pi}{8}. For k=2k=2, you get 9π8\frac{9\pi}{8}, which is already outside the interval. Thus the correct solution set is π8,5π8\frac{\pi}{8}, \frac{5\pi}{8}. The choice listing only π8\frac{\pi}{8} is incomplete: it captures the first solution but misses the second angle within the interval. The choice π4,5π4\frac{\pi}{4}, \frac{5\pi}{4} suggests solving tan⁡x=1\tan x=1 instead of tan⁡2x=1\tan 2x=1, and it also includes an angle outside [0,π)[0,\pi). The four-solution list π8,5π8,9π8,13π8\frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8} correctly shows the pattern over [0,2π)[0,2\pi), but the last two values are outside the required interval [0,π)[0,\pi). A good study habit: always divide by the coefficient of xx after setting up the general solution, then list every angle in the given interval before choosing your answer.

Question 4

Solve 2cos⁡2x+3cos⁡x+1=02\cos^2 x + 3\cos x + 1 = 0 over the interval [0,2π)[0, 2\pi).

  1. 2π3,4π3\frac{2\pi}{3}, \frac{4\pi}{3}
  2. π3,5π3\frac{\pi}{3}, \frac{5\pi}{3}
  3. π3,π,5π3\frac{\pi}{3}, \pi, \frac{5\pi}{3}
  4. 2π3,π,4π3\frac{2\pi}{3}, \pi, \frac{4\pi}{3} (correct answer)
Explanation: When you see a trigonometric equation that is quadratic in shape, treat it like an algebra problem in disguise. Let u=cos⁡xu = \cos x, so 2u2+3u+1=02u^2 + 3u + 1 = 0. Factoring gives (2u+1)(u+1)=0(2u+1)(u+1)=0, so u=−12u = -\frac12 or u=−1u = -1. That means you need to solve cos⁡x=−12\cos x = -\frac12 and cos⁡x=−1\cos x = -1 on [0,2π)[0,2\pi). Cosine equals −12-\frac12 at x=2π3x = \frac{2\pi}{3} and x=4π3x = \frac{4\pi}{3}, and cosine equals −1-1 at x=πx = \pi. Therefore the full solution set is {2π3,π,4π3}\left\{\frac{2\pi}{3}, \pi, \frac{4\pi}{3}\right\}. Now look at the traps. The choice listing only 2π3\frac{2\pi}{3} and 4π3\frac{4\pi}{3} correctly solves cos⁡x=−12\cos x = -\frac12 but forgets the second factor cos⁡x+1=0\cos x + 1 = 0, which gives x=πx = \pi. The choice with π3\frac{\pi}{3} and 5π3\frac{5\pi}{3} comes from mistakenly solving cos⁡x=12\cos x = \frac12 instead of −12-\frac12; those angles have positive cosine. The mixed choice π3,π,5π3\frac{\pi}{3}, \pi, \frac{5\pi}{3} correctly includes π\pi, but again uses the positive-half solutions instead of the negative-half solutions, so it misses the true angles. Every wrong option either drops a factor or flips the sign of cos⁡x\cos x. A strong habit: after factoring a trig quadratic, list the simple trig equations separately, solve each one over the given interval, then combine all solutions. Double-check cosine signs by thinking of the unit circle quadrants — negative cosine values live in Quadrants II and III.

Question 5

Solve 2sin⁡2x=1−cos⁡x2\sin^2 x = 1 - \cos x over the interval [0,2π)[0, 2\pi).

  1. π3,π,5π3\frac{\pi}{3}, \pi, \frac{5\pi}{3}
  2. 2π3,4π3\frac{2\pi}{3}, \frac{4\pi}{3}
  3. 0,2π3,4π30, \frac{2\pi}{3}, \frac{4\pi}{3} (correct answer)
  4. 0,π3,5π30, \frac{\pi}{3}, \frac{5\pi}{3}
Explanation: When you see sin⁡2x\sin^2 x and cos⁡x\cos x in the same equation, your first move should be to use the Pythagorean identity sin⁡2x=1−cos⁡2x\sin^2 x = 1-\cos^2 x so you have only one trig function. Substituting gives 2(1−cos⁡2x)=1−cos⁡x2(1-\cos^2 x)=1-\cos x which simplifies to 2cos⁡2x−cos⁡x−1=0.2\cos^2 x-\cos x-1=0. This factors as (2cos⁡x+1)(cos⁡x−1)=0,(2\cos x+1)(\cos x-1)=0, so cos⁡x=−12\cos x=-\frac12 or cos⁡x=1\cos x=1. Over [0,2π)[0,2\pi), cos⁡x=−12\cos x=-\frac12 at x=2π3x=\frac{2\pi}{3} and x=4π3x=\frac{4\pi}{3}, while cos⁡x=1\cos x=1 at x=0x=0. Thus the solution set is 0,2π3,4π30,\frac{2\pi}{3},\frac{4\pi}{3}. The choice π3,π,5π3\frac{\pi}{3},\pi,\frac{5\pi}{3} comes from a factoring sign error: using (2cos⁡x−1)(cos⁡x+1)=0(2\cos x-1)(\cos x+1)=0 would give cos⁡x=12\cos x=\frac12 and cos⁡x=−1\cos x=-1, which do not satisfy the original equation. The choice 2π3,4π3\frac{2\pi}{3},\frac{4\pi}{3} correctly handles the cosine-negative branch but forgets the cos⁡x=1\cos x=1 solution, so 00 is missing. The choice 0,π3,5π30,\frac{\pi}{3},\frac{5\pi}{3} keeps 00 but repeats the same sign error, swapping −12-\frac12 for 12\frac12. A quick habit that saves points: after solving a trig equation, plug your angles back into the original equation. Also, when factoring, check your signs carefully — one sign error changes the whole solution set.

Question 6

Solve 2sin⁡2x−sin⁡x−1=02\sin^2 x - \sin x - 1 = 0 over the interval [0,2π)[0, 2\pi).

  1. π2,3π2\frac{\pi}{2}, \frac{3\pi}{2}
  2. 7π6,11π6\frac{7\pi}{6}, \frac{11\pi}{6}
  3. π2,7π6,11π6\frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6} (correct answer)
  4. π6,5π6,3π2\frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}
Explanation: Whenever you see an equation like 2sin⁡2x−sin⁡x−1=02\sin^2 x-\sin x-1=0, recognize it as a quadratic in sin⁡x\sin x. Let u=sin⁡xu=\sin x; then 2u2−u−1=02u^2-u-1=0, which factors nicely: (2u+1)(u−1)=0(2u+1)(u-1)=0. So sin⁡x=1\sin x=1 or sin⁡x=−12\sin x=-\frac12. On [0,2π)[0,2\pi), sin⁡x=1\sin x=1 occurs only at x=π2x=\frac\pi2. For sin⁡x=−12\sin x=-\frac12, sine is negative in quadrants III andIV; the reference angle is π6\frac\pi6; this gives x=π+π6=7π6x=\pi+\frac\pi6=\frac{7\pi}{6} and x=2π−π6=11π6x=2\pi-\frac\pi6=\frac{11\pi}{6}. Therefore, the full solution set is π2,7π6,11π6\frac\pi2,\frac{7\pi}{6},\frac{11\pi}{6}. The choice containing π2\frac\pi2 and3π2\frac{3\pi}{2} include sth 3π/23\pi/2, where sin⁡x=−1\sin x=-1; plugging x=3π/2x=3\pi/2 into the original gives 2(1)−(−1)−1=2≠02(1)-(-1)-1=2\neq0, so it cannot be aroot! The one with only 7π6\frac{7\pi}{6} and11π6\frac{11\pi}{6} finds the negative-half branch but forgets the sin⁡x=1\sin x=1 branch. The choice with π6,5π6,3π2\frac\pi6,\frac{5\pi}{6},\frac{3\pi}{2} usesth e positive-half angles +12+\frac12, whenyou need the negative-half values, and also adds the invalid 3π/23\pi/2. For trig quadratics, factor first, then solve each simple equation. Draw a quick unit circle and check signs: sine negative means quadrants III andIV, notQI/II.

Question 7

Solve 2sin⁡x+2=02\sin x + \sqrt{2} = 0 over the interval [−π,π][-\pi, \pi].

  1. −π4-\frac{\pi}{4}
  2. −3π4,−π4-\frac{3\pi}{4}, -\frac{\pi}{4} (correct answer)
  3. −π4,3π4-\frac{\pi}{4}, \frac{3\pi}{4}
  4. 5π4,7π4\frac{5\pi}{4}, \frac{7\pi}{4}
Explanation: When you see a trigonometric equation over a restricted interval, your first move is to isolate the trig function and then use the unit circle to find every angle in that interval with the required sine value. Here, 2sin⁡x+2=02\sin x+\sqrt{2}= 0 becomes sin⁡x=−22\sin x = -\frac{\sqrt{2}}{2}. The reference angle for ∣sin⁡x∣=22|\sin x|=\frac{\sqrt{2}}{2} is π/4\pi/4. Since sine is negative, xx must lie inquadrants III or IV. Over [−π,π][-\pi,\pi], that means x=−3π4x = -\frac{3\pi}{4} (quadrant III, as a negative angle from −π-\pi) and x=−π4x = -\frac{\pi}{4} (quadrant IV). Checking: 2sin⁡(−3π/4)+2=2(−2/2)+2=02\sin(-3\pi/4)+\sqrt2=2(-\sqrt2/2)+\sqrt2=0, and the same for −π/4-\pi/4. So the full solution set is −3π4,−π4-\frac{3\pi}{4},-\frac{\pi}{4}. Looking at the temptations: the answer listing only −π4-\frac{\pi}{4} is incomplete — it recognizes quadrant IV but forgets the quadrant III solution withinthe interval. The pair −π4,3π4-\frac{\pi}{4},\frac{3\pi}{4} swaps in 3π/43\pi/4, but its sine is positive 22\frac{\sqrt2}{2}, so it doesn't satisfy the equation. The pair 5π4,7π4\frac{5\pi}{4},\frac{7\pi}{4} gives correct coterminal angles for the sign, but both are outside [−π,π][-\pi,\pi]; if the interval were restricted differently they might work, but here they must be excluded. Study tip: For trig equations, always sketch or visualize the four quadrants, then list only solutions inside the given interval. Coterminal angles are not automatically valid — check the bounds first.

Question 8

Solve cos⁡(2x)=−22\cos(2x) = -\frac{\sqrt{2}}{2} over the interval [0,2π)[0, 2\pi).

  1. 3π8,5π8\frac{3\pi}{8}, \frac{5\pi}{8}
  2. 3π8,5π8,11π8,13π8\frac{3\pi}{8}, \frac{5\pi}{8}, \frac{11\pi}{8}, \frac{13\pi}{8} (correct answer)
  3. 3π4,5π4\frac{3\pi}{4}, \frac{5\pi}{4}
  4. π8,7π8,9π8,15π8\frac{\pi}{8}, \frac{7\pi}{8}, \frac{9\pi}{8}, \frac{15\pi}{8}
Explanation: Whenever you're solving a trig equation with a doubled argument, the key is to let θ=2x\theta=2x and check the full range of θ\theta before dividing. Here cos⁡θ=−22\cos\theta=-\frac{\sqrt{2}}{2}. Since cosine is negative in quadrant II and III, the reference angle π4\frac{\pi}{4} gives θ=3π4,5π4\theta=\frac{3\pi}{4},\frac{5\pi}{4} within one full revolution. But xx runs from 00 to 2π2\pi, so θ=2x\theta=2x runs from 00 to 4π4\pi. You need every solution over two revolutions: add 2π2\pi to each base angle, obtaining θ=3π4,5π4,11π4,13π4\theta=\frac{3\pi}{4},\frac{5\pi}{4},\frac{11\pi}{4},\frac{13\pi}{4}. Dividing by 22 yields 3π8,5π8,11π8,13π8\frac{3\pi}{8},\frac{5\pi}{8},\frac{11\pi}{8},\frac{13\pi}{8}. The alternative with only 3π8,5π8\frac{3\pi}{8},\frac{5\pi}{8} simply forgets the second revolution of θ\theta, so it misses two solutions. The choice 3π4,5π4\frac{3\pi}{4},\frac{5\pi}{4} treats the equation as if it were cos⁡x=−22\cos x=-\frac{\sqrt{2}}{2}; it ignores the factor of 22 entirely, and those values actually make cos⁡(2x)=0\cos(2x)=0, not −22-\frac{\sqrt{2}}{2}. The list π8,7π8,9π8,15π8\frac{\pi}{8},\frac{7\pi}{8},\frac{9\pi}{8},\frac{15\pi}{8} is the solution to cos⁡(2x)=+22\cos(2x)=+\frac{\sqrt{2}}{2}: it uses angles where cosine is positive, so the sign of the given value was dropped. A good habit: when solving cos⁡(inside)=c\cos(\text{inside})=c, first solve for the inside angle over the entire possible interval, then divide—and check one solution in the original equation to catch sign errors.

Question 9

Solve sin⁡(2x)=cos⁡x\sin(2x) = \cos x over the interval [0,2π)[0, 2\pi).

  1. π6,5π6\frac{\pi}{6}, \frac{5\pi}{6}
  2. π2,3π2\frac{\pi}{2}, \frac{3\pi}{2}
  3. π6,π2,5π6,3π2\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}, \frac{3\pi}{2} (correct answer)
  4. π6,5π6,3π2\frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}
Explanation: When you see an equation mixing sin⁡(2x)\sin(2x) and cos⁡x\cos x, your first move should be to use the double-angle identity sin⁡(2x)=2sin⁡xcos⁡x\sin(2x)=2\sin x\cos x. That transforms the equation into 2sin⁡xcos⁡x=cos⁡x2\sin x\cos x=\cos x. Bring everything to one side: 2sin⁡xcos⁡x−cos⁡x=02\sin x\cos x-\cos x=0, then factor: cos⁡x(2sin⁡x−1)=0\cos x(2\sin x-1)=0. By the zero-product property, either cos⁡x=0\cos x=0 or 2sin⁡x−1=02\sin x-1=0. On [0,2π)[0,2\pi), cos⁡x=0\cos x=0 at x=π2,3π2x=\frac{\pi}{2},\frac{3\pi}{2}, and sin⁡x=12\sin x=\frac12 at x=π6,5π6x=\frac{\pi}{6},\frac{5\pi}{6}. So the full solution set is π6,π2,5π6,3π2\frac{\pi}{6},\frac{\pi}{2},\frac{5\pi}{6},\frac{3\pi}{2}. The choice π6,5π6\frac{\pi}{6},\frac{5\pi}{6} only handles the sin⁡x=12\sin x=\frac12 branch and drops the cos⁡x=0\cos x=0 factor. The choice π2,3π2\frac{\pi}{2},\frac{3\pi}{2} does the reverse: it includes only the cosine branch. The choice π6,5π6,3π2\frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{2} includes both sine solutions and one cosine zero, but omits π2\frac{\pi}{2}, so it is incomplete. Study tip: never divide both sides by cos⁡x\cos x or any factor that could be zero. Factoring, not dividing, preserves every solution. Then use the unit circle to list all angles for each factor over the required interval.

Question 10

Solve 2sin⁡2x−1=02\sin^2 x - 1 = 0 over the interval [0,2π)[0, 2\pi).

  1. π4,3π4\frac{\pi}{4}, \frac{3\pi}{4}
  2. π4,3π4,5π4\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}
  3. π4,5π4\frac{\pi}{4}, \frac{5\pi}{4}
  4. π4,3π4,5π4,7π4\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} (correct answer)
Explanation: When you see a squared trig equation, your first move should be to isolate the trig function. Here, adding 1 and dividing by 2 gives sin⁡2x=12\sin^2 x = \frac12, so sin⁡x=±22\sin x = \pm \frac{\sqrt2}{2}. The plus sign means xx could be π4\frac{\pi}{4} or 3π4\frac{3\pi}{4}, since sine is positive in quadrants I and II. The minus sign gives 5π4\frac{5\pi}{4} or 7π4\frac{7\pi}{4}, where sine is negative in quadrants III and IV. Therefore, the complete solution set is π4,3π4,5π4,7π4\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}. The choice that lists only π4\frac{\pi}{4} and 3π4\frac{3\pi}{4} ignores the negative square root entirely. The choice with π4,3π4,5π4\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4} stops after finding three angles and misses the fourth quadrant solution. The pair π4,5π4\frac{\pi}{4}, \frac{5\pi}{4} mistakes the symmetry, keeping just one positive and one negative solution instead of accounting for all four quadrant angles where ∣sin⁡x∣=22|\sin x| = \frac{\sqrt2}{2}. A good habit: whenever you take a square root of a trig equation, write the ±\pm explicitly. Then use the unit circle to list every angle in the interval where sine has that sign. For a full period, you should expect a balanced set of quadrant answers—missing one usually means you forgot the sign or skipped a quadrant.

Question 11

Solve csc⁡x=2\csc x = 2 over the interval [0,2π)[0, 2\pi).

  1. π6,5π6\frac{\pi}{6}, \frac{5\pi}{6} (correct answer)
  2. π3,5π3\frac{\pi}{3}, \frac{5\pi}{3}
  3. π6\frac{\pi}{6}
  4. π6,5π6,7π6,11π6\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}
Explanation: When you see csc⁡x\csc x, think "reciprocal of sine." So csc⁡x=2\csc x = 2 means sin⁡x=12\sin x = \frac{1}{2}. Now you're solving a familiar sine equation on [0,2π)[0,2\pi). Sine is positive in Quadrants I and II, and the reference angle for sin⁡x=12\sin x=\frac12 is π6\frac{\pi}{6}. That gives two solutions: π6\frac{\pi}{6} in Quadrant I and 5π6\frac{5\pi}{6} in Quadrant II. Both check: csc⁡π6=2\csc\frac{\pi}{6}=2 and csc⁡5π6=2\csc\frac{5\pi}{6}=2. The choice π6\frac{\pi}{6} alone is incomplete—it captures only the Quadrant I answer. The choice π3,5π3\frac{\pi}{3}, \frac{5\pi}{3} is a classic trap: those are the solutions to cos⁡x=12\cos x=\frac12, not sin⁡x=12\sin x=\frac12. The four-angle choice π6,5π6,7π6,11π6\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6} adds two angles in Quadrants III and IV, where sine is negative, so csc⁡x\csc x would be negative, never 22. A quick strategy: always convert reciprocal trig functions to their basic form first. Then ask yourself where the basic function has the required sign—for sine, that's QI and QII—and use the reference angle in each relevant quadrant. This prevents both the "only one angle" mistake and the "extra quadrants" mistake.

Question 12

Solve 2cos⁡2x−3cos⁡x+1=02\cos^2 x - 3\cos x + 1 = 0 over the interval [0,2π)[0, 2\pi).

  1. 0,π3,5π30, \frac{\pi}{3}, \frac{5\pi}{3} (correct answer)
  2. π3,5π3\frac{\pi}{3}, \frac{5\pi}{3}
  3. 0,2π3,4π30, \frac{2\pi}{3}, \frac{4\pi}{3}
  4. π3,π,5π3\frac{\pi}{3}, \pi, \frac{5\pi}{3}
Explanation: Whenever you see a quadratic expression involving a trig function, treat it like a standard quadratic. Let u=cos⁡xu = \cos x — the equation becomes 2u2−3u+1=02u^2 - 3u + 1 = 0, which factors cleanly. Factoring gives (2u−1)(u−1)=0(2u - 1)(u - 1) = 0, so 2cos⁡x−1=02\cos x - 1 = 0 or cos⁡x−1=0\cos x - 1 = 0. This yields cos⁡x=12\cos x = \frac{1}{2} or cos⁡x=1\cos x = 1. Over [0,2π)[0, 2\pi), cos⁡x=1\cos x = 1 at x=0x = 0. For cos⁡x=12\cos x = \frac{1}{2}, cosine is positive in quadrants I and IV, giving x=π3x = \frac{\pi}{3} and x=5π3x = \frac{5\pi}{3}. Thus the solution set is {0,π3,5π3}\{0, \frac{\pi}{3}, \frac{5\pi}{3}\}. Now look at the other choices. The set {π3,5π3}\{\frac{\pi}{3}, \frac{5\pi}{3}\} is missing 00 — that trap appears if you forget that cos⁡x=1\cos x = 1 has a solution at the origin. The set {0,2π3,4π3}\{0, \frac{2\pi}{3}, \frac{4\pi}{3}\} uses angles where cos⁡x=−12\cos x = -\frac{1}{2}, not +12+\frac{1}{2}; this is a sign error on the unit circle (remember cosine is positive in QI and QIV). The set {π3,π,5π3}\{\frac{\pi}{3}, \pi, \frac{5\pi}{3}\} incorrectly includes π\pi, which would require cos⁡x=−1\cos x = -1. That mistake comes from mis-factoring the quadratic, perhaps as (2cos⁡x−1)(cos⁡x+1)=0(2\cos x - 1)(\cos x + 1) = 0. Your takeaway: when solving a trig quadratic, always solve for each factor separately, list every solution in the given interval, and double-check your unit circle values — especially the signs of cosine in each quadrant.