Solving Trigonometric Equations - Algebra 3

1. How many solutions does $\sin 2x = \cos x$ have on the interval $[0,2\pi)$?
   Answer: 4 Rewrite $\sin 2x$ as $2\sin x\cos x$, so $2\sin x\cos x=\cos x$. Then $\cos x(2\sin x-1)=0$. On $[0,2\pi)$, $\cos x=0$ gives $\pi/2$ and $3\pi/2$, while $\sin x=1/2$ gives $\pi/6$ and $5\pi/6$. That is 4 solutions. Counting only one factor, or incorrectly adding an endpoint solution, gives the other counts.
2. Solve $\cos 2x = \cos x$ on the interval $[0,2\pi)$.
   Answer: $\{0,\frac{2\pi}{3},\frac{4\pi}{3}\}$ Let $c=\cos x$. Since $\cos 2x=2c^2-1$, the equation becomes $2c^2-1=c$, so $2c^2-c-1=0$, or $(2c+1)(c-1)=0$. Thus $\cos x=-1/2$, giving $2\pi/3$ and $4\pi/3$, or $\cos x=1$, giving $0$. The set with $\pi/3$ and $5\pi/3$ comes from using $\cos x=1/2$ instead of $-1/2$; the set with $\pi/2$ and $3\pi/2$ incorrectly includes values where $\cos x=0$; the set with only $2\pi/3$ and $4\pi/3$ omits the solution $0$.
3. Solve $2\sin^2 x = 1 - \cos x$ over the interval $[0, 2\pi)$.
   Answer: $0, \frac{2\pi}{3}, \frac{4\pi}{3}$ When you see $\sin^2 x$ and $\cos x$ in the same equation, your first move should be to use the Pythagorean identity $\sin^2 x = 1-\cos^2 x$ so you have only one trig function. Substituting gives $$2(1-\cos^2 x)=1-\cos x$$ which simplifies to $$2\cos^2 x-\cos x-1=0.$$ This factors as $$(2\cos x+1)(\cos x-1)=0,$$ so $\cos x=-\frac12$ or $\cos x=1$. Over $[0,2\pi)$, $\cos x=-\frac12$ at $x=\frac{2\pi}{3}$ and $x=\frac{4\pi}{3}$, while $\cos x=1$ at $x=0$. Thus the solution set is $0,\frac{2\pi}{3},\frac{4\pi}{3}$. The choice $\frac{\pi}{3},\pi,\frac{5\pi}{3}$ comes from a factoring sign error: using $(2\cos x-1)(\cos x+1)=0$ would give $\cos x=\frac12$ and $\cos x=-1$, which do not satisfy the original equation. The choice $\frac{2\pi}{3},\frac{4\pi}{3}$ correctly handles the cosine-negative branch but forgets the $\cos x=1$ solution, so $0$ is missing. The choice $0,\frac{\pi}{3},\frac{5\pi}{3}$ keeps $0$ but repeats the same sign error, swapping $-\frac12$ for $\frac12$. 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.
4. Solve $2\sin^2 x - \sin x - 1 = 0$ over the interval $[0, 2\pi)$.
   Answer: $\frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}$ Whenever you see an equation like $2\sin^2 x-\sin x-1=0$, recognize it as a quadratic in $\sin x$. Let $u=\sin x$; then $2u^2-u-1=0$, which factors nicely: $(2u+1)(u-1)=0$. So $\sin x=1$ or $\sin x=-\frac12$. On $[0,2\pi)$, $\sin x=1$ occurs only at $x=\frac\pi2$. For $\sin x=-\frac12$, sine is negative in quadrants III andIV; the reference angle is $\frac\pi6$; this gives $x=\pi+\frac\pi6=\frac{7\pi}{6}$ and $x=2\pi-\frac\pi6=\frac{11\pi}{6}$. Therefore, the full solution set is $\frac\pi2,\frac{7\pi}{6},\frac{11\pi}{6}$. The choice containing $\frac\pi2$ and$\frac{3\pi}{2}$ include sth $3\pi/2$, where $\sin x=-1$; plugging $x=3\pi/2$ into the original gives $2(1)-(-1)-1=2\neq0$, so it cannot be aroot! The one with only $\frac{7\pi}{6}$ and$\frac{11\pi}{6}$ finds the negative-half branch but forgets the $\sin x=1$ branch. The choice with $\frac\pi6,\frac{5\pi}{6},\frac{3\pi}{2}$ usesth e positive-half angles $+\frac12$, whenyou need the negative-half values, and also adds the invalid $3\pi/2$. 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.
5. Solve $\sqrt{2}\sin x - \sqrt{2}\cos x = 2$ on the interval $[0,2\pi)$.
   Answer: $\{\frac{3\pi}{4}\}$ Divide by $\sqrt{2}$: $\sin x-\cos x=\sqrt{2}$. Squaring gives $\sin^2x-2\sin x\cos x+\cos^2x=2$, so $1-\sin 2x=2$ and $\sin 2x=-1$. On $[0,2\pi)$, this gives candidate solutions $x=3\pi/4$ and $x=7\pi/4$. Substituting $3\pi/4$ checks, but $7\pi/4$ gives $-2$, not $2$, so it is extraneous. The only solution is $3\pi/4$.
6. Solve $\csc x = 2$ over the interval $[0, 2\pi)$.
   Answer: $\frac{\pi}{6}, \frac{5\pi}{6}$ When you see $\csc x$, think "reciprocal of sine." So $\csc x = 2$ means $\sin x = \frac{1}{2}$. Now you're solving a familiar sine equation on $[0,2\pi)$. Sine is positive in Quadrants I and II, and the reference angle for $\sin x=\frac12$ is $\frac{\pi}{6}$. That gives two solutions: $\frac{\pi}{6}$ in Quadrant I and $\frac{5\pi}{6}$ in Quadrant II. Both check: $\csc\frac{\pi}{6}=2$ and $\csc\frac{5\pi}{6}=2$. The choice $\frac{\pi}{6}$ alone is incomplete—it captures only the Quadrant I answer. The choice $\frac{\pi}{3}, \frac{5\pi}{3}$ is a classic trap: those are the solutions to $\cos x=\frac12$, not $\sin x=\frac12$. The four-angle choice $\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$ adds two angles in Quadrants III and IV, where sine is negative, so $\csc x$ would be negative, never $2$. 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.
7. Solve $2\sin^2 x - 1 = 0$ over the interval $[0, 2\pi)$.
   Answer: $\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ 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 $\sin^2 x = \frac12$, so $\sin x = \pm \frac{\sqrt2}{2}$. The plus sign means $x$ could be $\frac{\pi}{4}$ or $\frac{3\pi}{4}$, since sine is positive in quadrants I and II. The minus sign gives $\frac{5\pi}{4}$ or $\frac{7\pi}{4}$, where sine is negative in quadrants III and IV. Therefore, the complete solution set is $\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$. The choice that lists only $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ ignores the negative square root entirely. The choice with $\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}$ stops after finding three angles and misses the fourth quadrant solution. The pair $\frac{\pi}{4}, \frac{5\pi}{4}$ mistakes the symmetry, keeping just one positive and one negative solution instead of accounting for all four quadrant angles where $|\sin x| = \frac{\sqrt2}{2}$. 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.
8. Solve $2\cos^2 x + 3\cos x + 1 = 0$ over the interval $[0, 2\pi)$.
   Answer: $\frac{2\pi}{3}, \pi, \frac{4\pi}{3}$ When you see a trigonometric equation that is quadratic in shape, treat it like an algebra problem in disguise. Let $u = \cos x$, so $2u^2 + 3u + 1 = 0$. Factoring gives $(2u+1)(u+1)=0$, so $u = -\frac12$ or $u = -1$. That means you need to solve $\cos x = -\frac12$ and $\cos x = -1$ on $[0,2\pi)$. Cosine equals $-\frac12$ at $x = \frac{2\pi}{3}$ and $x = \frac{4\pi}{3}$, and cosine equals $-1$ at $x = \pi$. Therefore the full solution set is $\left\{\frac{2\pi}{3}, \pi, \frac{4\pi}{3}\right\}$. Now look at the traps. The choice listing only $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$ correctly solves $\cos x = -\frac12$ but forgets the second factor $\cos x + 1 = 0$, which gives $x = \pi$. The choice with $\frac{\pi}{3}$ and $\frac{5\pi}{3}$ comes from mistakenly solving $\cos x = \frac12$ instead of $-\frac12$; those angles have positive cosine. The mixed choice $\frac{\pi}{3}, \pi, \frac{5\pi}{3}$ correctly includes $\pi$, 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 $\cos x$. 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.
9. Solve $\tan(2x) = 1$ over the interval $[0, \pi)$.
   Answer: $\frac{\pi}{8}, \frac{5\pi}{8}$ Whenever you see a trigonometric equation like $\tan(2x)=1$, remember that the tangent function has period $\pi$, not $2\pi$. So $\tan(\theta)=1$ when $$\theta=\frac{\pi}{4}+k\pi$$ for any integer $k$. Here $\theta=2x$, so set $$2x=\frac{\pi}{4}+k\pi \quad\Rightarrow\quad x=\frac{\pi}{8}+\frac{k\pi}{2}.$$ Now find which of these values lie in $[0,\pi)$. For $k=0$, you get $\frac{\pi}{8}$. For $k=1$, you get $\frac{\pi}{8}+\frac{\pi}{2}=\frac{5\pi}{8}$. For $k=2$, you get $\frac{9\pi}{8}$, which is already outside the interval. Thus the correct solution set is $\frac{\pi}{8}, \frac{5\pi}{8}$. The choice listing only $\frac{\pi}{8}$ is incomplete: it captures the first solution but misses the second angle within the interval. The choice $\frac{\pi}{4}, \frac{5\pi}{4}$ suggests solving $\tan x=1$ instead of $\tan 2x=1$, and it also includes an angle outside $[0,\pi)$. The four-solution list $\frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8}$ correctly shows the pattern over $[0,2\pi)$, but the last two values are outside the required interval $[0,\pi)$. A good study habit: always divide by the coefficient of $x$ after setting up the general solution, then list every angle in the given interval before choosing your answer.
10. Solve $\tan^2 x - 3\tan x + 2 = 0$ on the interval $[0,2\pi)$.
    Answer: $\{\frac{\pi}{4},\arctan 2,\frac{5\pi}{4},\pi+\arctan 2\}$ Let $t=\tan x$. Then $t^2-3t+2=(t-1)(t-2)$, so $\tan x=1$ or $\tan x=2$. $\tan x=1$ gives $\pi/4$ and $5\pi/4$. Since tangent has period $\pi$, $\tan x=2$ gives $\arctan 2$ and $\pi+\arctan 2$. A set containing only $\arctan 2$ or only $\pi/4,5\pi/4$ is missing one factor; a set without $\pi+\arctan 2$ has not added the period for the second tangent value.
11. Solve $\tan^2 x + \sec x = 1$ on the interval $[0,2\pi)$.
    Answer: $\{0,\frac{2\pi}{3},\frac{4\pi}{3}\}$ Use $\tan^2 x=\sec^2 x-1$. The equation becomes $\sec^2 x+\sec x-2=0$, or $(\sec x+2)(\sec x-1)=0$. $\sec x=-2$ gives $\cos x=-1/2$, so $2\pi/3$ and $4\pi/3$. $\sec x=1$ gives $\cos x=1$, so $0$. The sets involving $\pi/3$ and $5\pi/3$ usually come from solving $\cos x=1/2$ instead of $-1/2$ or from an incorrect factorization.
12. Solve $\sin x + \sin 2x = 0$ on the interval $[0,2\pi)$.
    Answer: $\{0,\frac{2\pi}{3},\pi,\frac{4\pi}{3}\}$ Since $\sin 2x=2\sin x\cos x$, the equation becomes $\sin x(1+2\cos x)=0$. $\sin x=0$ gives $0$ and $\pi$. Also, $1+2\cos x=0$ gives $\cos x=-1/2$, so $2\pi/3$ and $4\pi/3$. The complete set is $0,\ 2\pi/3,\ \pi,\ 4\pi/3$. Omitting the $\sin x=0$ roots leaves only $2\pi/3$ and $4\pi/3$; replacing $1+2\cos x$ with $1-2\cos x$ gives $\pi/3$ and $5\pi/3$.
13. Solve $\cos(\pi x) = \frac{\sqrt{2}}{2}$ over the interval $0 \le x \le 2$.
    Answer: $\frac{1}{4}, \frac{7}{4}$ When you solve $\cos(\pi x)=\frac{\sqrt2}{2}$, you're really solving the general cosine equation $\cos \theta=\frac{\sqrt2}{2}$ with $\theta=\pi x$. Cosine is positive in Quadrants I and IV, so the reference angle $\frac{\pi}{4}$ gives $\theta=\frac{\pi}{4}$ and $\theta=\frac{7\pi}{4}$ after adding multiples of $2\pi$. Dividing by $\pi$, this becomes $x=\frac14+2k$ or $x=\frac74+2k$. On $0\le x\le 2$, only $x=\frac14$ and $x=\frac74$ fit, so the correct answer is $\frac14,\frac74$. The single answer $\frac14$ misses the second valid solution in Quadrant IV. The answer $\frac14,\frac74,\frac94,\frac{15}{4}$ correctly identifies the pattern but includes $\frac94$ and $\frac{15}{4}$, which are outside the interval $0$ to $2$. The answer $\frac14,\frac34,\frac54,\frac74$ confuses angles where cosine equals $\pm\frac{\sqrt2}{2}$: $\frac{3\pi}{4}$ and $\frac{5\pi}{4}$ 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.
14. Solve $\cos(2x) = -\frac{\sqrt{2}}{2}$ over the interval $[0, 2\pi)$.
    Answer: $\frac{3\pi}{8}, \frac{5\pi}{8}, \frac{11\pi}{8}, \frac{13\pi}{8}$ Whenever you're solving a trig equation with a doubled argument, the key is to let $\theta=2x$ and check the full range of $\theta$ before dividing. Here $\cos\theta=-\frac{\sqrt{2}}{2}$. Since cosine is negative in quadrant II and III, the reference angle $\frac{\pi}{4}$ gives $\theta=\frac{3\pi}{4},\frac{5\pi}{4}$ within one full revolution. But $x$ runs from $0$ to $2\pi$, so $\theta=2x$ runs from $0$ to $4\pi$. You need every solution over two revolutions: add $2\pi$ to each base angle, obtaining $\theta=\frac{3\pi}{4},\frac{5\pi}{4},\frac{11\pi}{4},\frac{13\pi}{4}$. Dividing by $2$ yields $\frac{3\pi}{8},\frac{5\pi}{8},\frac{11\pi}{8},\frac{13\pi}{8}$. The alternative with only $\frac{3\pi}{8},\frac{5\pi}{8}$ simply forgets the second revolution of $\theta$, so it misses two solutions. The choice $\frac{3\pi}{4},\frac{5\pi}{4}$ treats the equation as if it were $\cos x=-\frac{\sqrt{2}}{2}$; it ignores the factor of $2$ entirely, and those values actually make $\cos(2x)=0$, not $-\frac{\sqrt{2}}{2}$. The list $\frac{\pi}{8},\frac{7\pi}{8},\frac{9\pi}{8},\frac{15\pi}{8}$ is the solution to $\cos(2x)=+\frac{\sqrt{2}}{2}$: it uses angles where cosine is positive, so the sign of the given value was dropped. A good habit: when solving $\cos(\text{inside})=c$, 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.
15. Solve $2\sin x + \sqrt{2} = 0$ over the interval $[-\pi, \pi]$.
    Answer: $-\frac{3\pi}{4}, -\frac{\pi}{4}$ 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, $2\sin x+\sqrt{2}= 0$ becomes $\sin x = -\frac{\sqrt{2}}{2}$. The reference angle for $|\sin x|=\frac{\sqrt{2}}{2}$ is $\pi/4$. Since sine is negative, $x$ must lie inquadrants III or IV. Over $[-\pi,\pi]$, that means $x = -\frac{3\pi}{4}$ (quadrant III, as a negative angle from $-\pi$) and $x = -\frac{\pi}{4}$ (quadrant IV). Checking: $2\sin(-3\pi/4)+\sqrt2=2(-\sqrt2/2)+\sqrt2=0$, and the same for $-\pi/4$. So the full solution set is $-\frac{3\pi}{4},-\frac{\pi}{4}$. Looking at the temptations: the answer listing only $-\frac{\pi}{4}$ is incomplete — it recognizes quadrant IV but forgets the quadrant III solution withinthe interval. The pair $-\frac{\pi}{4},\frac{3\pi}{4}$ swaps in $3\pi/4$, but its sine is positive $\frac{\sqrt2}{2}$, so it doesn't satisfy the equation. The pair $\frac{5\pi}{4},\frac{7\pi}{4}$ gives correct coterminal angles for the sign, but both are outside $[-\pi,\pi]$; 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.
16. Solve $2\cos^2 x - 3\cos x + 1 = 0$ over the interval $[0, 2\pi)$.
    Answer: $0, \frac{\pi}{3}, \frac{5\pi}{3}$ Whenever you see a quadratic expression involving a trig function, treat it like a standard quadratic. Let $u = \cos x$ — the equation becomes $2u^2 - 3u + 1 = 0$, which factors cleanly. Factoring gives $(2u - 1)(u - 1) = 0$, so $2\cos x - 1 = 0$ or $\cos x - 1 = 0$. This yields $\cos x = \frac{1}{2}$ or $\cos x = 1$. Over $[0, 2\pi)$, $\cos x = 1$ at $x = 0$. For $\cos x = \frac{1}{2}$, cosine is positive in quadrants I and IV, giving $x = \frac{\pi}{3}$ and $x = \frac{5\pi}{3}$. Thus the solution set is $\{0, \frac{\pi}{3}, \frac{5\pi}{3}\}$. Now look at the other choices. The set $\{\frac{\pi}{3}, \frac{5\pi}{3}\}$ is missing $0$ — that trap appears if you forget that $\cos x = 1$ has a solution at the origin. The set $\{0, \frac{2\pi}{3}, \frac{4\pi}{3}\}$ uses angles where $\cos x = -\frac{1}{2}$, not $+\frac{1}{2}$; this is a sign error on the unit circle (remember cosine is positive in QI and QIV). The set $\{\frac{\pi}{3}, \pi, \frac{5\pi}{3}\}$ incorrectly includes $\pi$, which would require $\cos x = -1$. That mistake comes from mis-factoring the quadratic, perhaps as $(2\cos x - 1)(\cos x + 1) = 0$. 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.
17. Which equation has exactly three solutions on $[0,2\pi)$?
    Answer: $\sin x(\sin x-1)=0$ Check each product. The product $\cos x(\sin x-1)=0$ has two solutions because $\cos x=0$ and $\sin x=1$ overlap at $\pi/2$. The product $\tan x(\sin x-1)=0$ also has two: $\tan x=0$ gives $0,\pi$, while $\sin x=1$ is not valid because $\tan x$ is undefined at $\pi/2$. The product $\sin x(\cos x-1)=0$ has two because $\sin x=0$ gives $0,\pi$ and $\cos x=1$ at $0$ is already counted. Only $\sin x(\sin x-1)=0$ gives three distinct solutions: $0,\pi/2,\pi$.
18. Solve $\sin x = \cos 2x$ on the interval $[0,2\pi)$.
    Answer: $\{\frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{2}\}$ Use $\cos 2x=1-2\sin^2 x$. If $s=\sin x$, then $1-2s^2=s$, so $2s^2+s-1=0$, or $(2s-1)(s+1)=0$. Thus $\sin x=1/2$, giving $\pi/6$ and $5\pi/6$, or $\sin x=-1$, giving $3\pi/2$. The choices missing $3\pi/2$ or $5\pi/6$ omit one of the factor or quadrant solutions.
19. Solve $\tan x \sin x = \tan x$ on the interval $[0,2\pi)$.
    Answer: Exactly two solutions: $0$ and $\pi$. Do not divide by $\tan x$. Factoring gives $\tan x(\sin x-1)=0$. $\tan x=0$ gives $0$ and $\pi$. Although $\sin x-1=0$ gives $\pi/2$, $\tan x$ is undefined there, so it is not a solution. Thus the only solutions are $0$ and $\pi$. The other choices include $\pi/2$ or $3\pi/2$ by dividing by $\tan x$ or by ignoring the domain restriction.
20. How many solutions does $2\sin^2 x = 1$ have on the interval $[0,3\pi]$?
    Answer: 6 $\sin^2 x=1/2$, so $\sin x=\pm\sqrt{2}/2$. On $[0,3\pi]$, $\sin x=\sqrt{2}/2$ at $\pi/4,\ 3\pi/4,\ 9\pi/4,\ 11\pi/4$, and $\sin x=-\sqrt{2}/2$ at $5\pi/4$ and $7\pi/4$. That is a total of 6 solutions. Counting only one full period gives 4; adding an endpoint solution or one extra period gives 7 or 8.

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