Math 3 Quiz: Solving Trig Equations
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Solving Trig EquationsQuestion 1 of 13

If tan⁡(θ)=−34\tan(\theta) = -\frac{3}{4} and θ\theta is in the second quadrant, what is the value of sin⁡(2θ)\sin(2\theta)?

−2425-\frac{24}{25}
−725-\frac{7}{25}
725\frac{7}{25}
2425\frac{24}{25}
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Math 3 Quiz

Math 3 Quiz: Solving Trig Equations

Practice Solving Trig Equations in Math 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 Trig Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 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

If tan⁡(θ)=−34\tan(\theta) = -\frac{3}{4} and θ\theta is in the second quadrant, what is the value of sin⁡(2θ)\sin(2\theta)?

  1. −2425-\frac{24}{25} (correct answer)
  2. −725-\frac{7}{25}
  3. 725\frac{7}{25}
  4. 2425\frac{24}{25}
Explanation: Since tan⁡(θ)=−34\tan(\theta) = -\frac{3}{4} and θ\theta is in quadrant II, we have sin⁡(θ)>0\sin(\theta) > 0 and cos⁡(θ)<0\cos(\theta) < 0. Using tan⁡(θ)=sin⁡(θ)cos⁡(θ)=−34\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = -\frac{3}{4}, we can write sin⁡(θ)=35\sin(\theta) = \frac{3}{5} and cos⁡(θ)=−45\cos(\theta) = -\frac{4}{5} (using the 3-4-5 triangle). Then sin⁡(2θ)=2sin⁡(θ)cos⁡(θ)=2⋅35⋅(−45)=−2425\sin(2\theta) = 2\sin(\theta)\cos(\theta) = 2 \cdot \frac{3}{5} \cdot (-\frac{4}{5}) = -\frac{24}{25}. Choice B results from using cos⁡(2θ)\cos(2\theta) formula instead. Choice C forgets the negative sign. Choice D uses wrong quadrant signs.

Question 2

The equation sin⁡(x)cos⁡(x)=34\sin(x) \cos(x) = \frac{\sqrt{3}}{4} has how many solutions in [0,2π)[0, 2\pi)?

  1. 0 solutions
  2. 2 solutions
  3. 4 solutions (correct answer)
  4. 6 solutions
Explanation: Using the double angle identity sin⁡(2x)=2sin⁡(x)cos⁡(x)\sin(2x) = 2\sin(x)\cos(x), we get 12sin⁡(2x)=34\frac{1}{2}\sin(2x) = \frac{\sqrt{3}}{4}, so sin⁡(2x)=32\sin(2x) = \frac{\sqrt{3}}{2}. The general solution is 2x=π3+2πk2x = \frac{\pi}{3} + 2\pi k or 2x=2π3+2πk2x = \frac{2\pi}{3} + 2\pi k for integer kk. This gives x=π6+πkx = \frac{\pi}{6} + \pi k or x=π3+πkx = \frac{\pi}{3} + \pi k. For x∈[0,2π)x \in [0, 2\pi): From x=π6+πkx = \frac{\pi}{6} + \pi k: x=π6,7π6x = \frac{\pi}{6}, \frac{7\pi}{6}. From x=π3+πkx = \frac{\pi}{3} + \pi k: x=π3,4π3x = \frac{\pi}{3}, \frac{4\pi}{3}. So we have 4 solutions total. Choice A suggests no solutions exist. Choice B misses half the solutions. Choice D over-counts.

Question 3

If sin⁡(3α)=0\sin(3\alpha) = 0 and 0≤α<2π30 \leq \alpha < \frac{2\pi}{3}, how many values of α\alpha satisfy this condition?

  1. 1 value
  2. 2 values (correct answer)
  3. 3 values
  4. 4 values
Explanation: sin⁡(3α)=0\sin(3\alpha) = 0 when 3α=nπ3\alpha = n\pi for integer nn, so α=nπ3\alpha = \frac{n\pi}{3}. For 0≤α<2π30 \leq \alpha < \frac{2\pi}{3}: when n=0n = 0: α=0\alpha = 0. When n=1n = 1: α=π3\alpha = \frac{\pi}{3}. When n=2n = 2: α=2π3\alpha = \frac{2\pi}{3}, but this equals the upper bound, so it's excluded since we have α<2π3\alpha < \frac{2\pi}{3}. Therefore, α=0,π3\alpha = 0, \frac{\pi}{3} are the only two solutions. Choice A misses one solution. Choice C incorrectly includes 2π3\frac{2\pi}{3}. Choice D over-counts significantly.

Question 4

For 0≤x≤π0 \leq x \leq \pi, the equation tan⁡2(x)−3tan⁡(x)+2=0\tan^2(x) - 3\tan(x) + 2 = 0 has how many solutions?

  1. 1 solution
  2. 2 solutions (correct answer)
  3. 3 solutions
  4. 4 solutions
Explanation: Let u=tan⁡(x)u = \tan(x). Then u2−3u+2=0u^2 - 3u + 2 = 0. Factoring: (u−1)(u−2)=0(u-1)(u-2) = 0, so u=1u = 1 or u=2u = 2. When tan⁡(x)=1\tan(x) = 1: x=π4x = \frac{\pi}{4} (in [0,π][0,\pi]). When tan⁡(x)=2\tan(x) = 2: x=arctan⁡(2)≈1.107x = \arctan(2) \approx 1.107 radians (in [0,π][0,\pi]). Note that tan⁡(x)\tan(x) is undefined at x=π2x = \frac{\pi}{2}, but this doesn't affect our solutions. Both solutions are in the given interval, so there are 2 solutions. Choice A misses one solution. Choice C might incorrectly include x=π+arctan⁡(2)x = \pi + \arctan(2), but this is outside our interval. Choice D over-counts.

Question 5

The equation cos⁡(x)−sin⁡(x)=2cos⁡(x+π4)\cos(x) - \sin(x) = \sqrt{2}\cos(x + \frac{\pi}{4}) is satisfied by:

  1. All real values of xx (correct answer)
  2. No real values of xx
  3. x=2πkx = 2\pi k for integer kk
  4. x=π+2πkx = \pi + 2\pi k for integer kk
Explanation: Using the cosine addition formula: cos⁡(x+π4)=cos⁡(x)cos⁡(π4)−sin⁡(x)sin⁡(π4)=cos⁡(x)⋅22−sin⁡(x)⋅22=22(cos⁡(x)−sin⁡(x))\cos(x + \frac{\pi}{4}) = \cos(x)\cos(\frac{\pi}{4}) - \sin(x)\sin(\frac{\pi}{4}) = \cos(x) \cdot \frac{\sqrt{2}}{2} - \sin(x) \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}(\cos(x) - \sin(x)). So the right side becomes: 2⋅22(cos⁡(x)−sin⁡(x))=cos⁡(x)−sin⁡(x)\sqrt{2} \cdot \frac{\sqrt{2}}{2}(\cos(x) - \sin(x)) = \cos(x) - \sin(x). The equation becomes cos⁡(x)−sin⁡(x)=cos⁡(x)−sin⁡(x)\cos(x) - \sin(x) = \cos(x) - \sin(x), which is an identity true for all real xx. Students might incorrectly try to solve this as a conditional equation rather than recognizing it as an identity.

Question 6

Which value of kk makes the equation sin⁡(x)+kcos⁡(x)=2\sin(x) + k\cos(x) = 2 have exactly one solution in [0,2π)[0, 2\pi)?

  1. k=3k = \sqrt{3} (correct answer)
  2. k=5k = \sqrt{5}
  3. k=2k = 2
  4. k=3k = 3
Explanation: The equation sin⁡(x)+kcos⁡(x)=2\sin(x) + k\cos(x) = 2 can be written as 1+k2sin⁡(x+ϕ)=2\sqrt{1+k^2}\sin(x + \phi) = 2 where tan⁡(ϕ)=k\tan(\phi) = k. This gives sin⁡(x+ϕ)=21+k2\sin(x + \phi) = \frac{2}{\sqrt{1+k^2}}. For exactly one solution in [0,2π)[0, 2\pi), we need the right side to equal 1 (maximum of sine), so 21+k2=1\frac{2}{\sqrt{1+k^2}} = 1. Solving: 2=1+k2⇒4=1+k2⇒k2=3⇒k=32 = \sqrt{1+k^2} \Rightarrow 4 = 1+k^2 \Rightarrow k^2 = 3 \Rightarrow k = \sqrt{3} (taking positive value). Choice B gives sin⁡(x+ϕ)=26<1\sin(x+\phi) = \frac{2}{\sqrt{6}} < 1, yielding two solutions. Choice C gives sin⁡(x+ϕ)=25<1\sin(x+\phi) = \frac{2}{\sqrt{5}} < 1, yielding two solutions. Choice D gives sin⁡(x+ϕ)=210<1\sin(x+\phi) = \frac{2}{\sqrt{10}} < 1, yielding two solutions.

Question 7

How many solutions does 2sin⁡2(x)+sin⁡(x)−1=02\sin^2(x) + \sin(x) - 1 = 0 have in the interval [0,2π)[0, 2\pi)?

  1. 1 solution
  2. 2 solutions
  3. 3 solutions (correct answer)
  4. 4 solutions
Explanation: Let u=sin⁡(x)u = \sin(x). Then 2u2+u−1=02u^2 + u - 1 = 0. Using the quadratic formula: u=−1±1+84=−1±34u = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}. So u=12u = \frac{1}{2} or u=−1u = -1. When sin⁡(x)=12\sin(x) = \frac{1}{2}: x=π6,5π6x = \frac{\pi}{6}, \frac{5\pi}{6} in [0,2π)[0, 2\pi). When sin⁡(x)=−1\sin(x) = -1: x=3π2x = \frac{3\pi}{2} in [0,2π)[0, 2\pi). Therefore, there are 3 solutions total: x=π6,5π6,3π2x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2}. Choice A counts only one of the sin⁡(x)=12\sin(x) = \frac{1}{2} solutions. Choice B misses the sin⁡(x)=−1\sin(x) = -1 solution. Choice D over-counts, perhaps by including extraneous solutions.

Question 8

For 0≤x≤π0 \leq x \leq \pi, the equation cos⁡(3x)=cos⁡(x)\cos(3x) = \cos(x) has how many solutions?

  1. 2 solutions
  2. 3 solutions (correct answer)
  3. 4 solutions
  4. 5 solutions
Explanation: The general solution to cos⁡(A)=cos⁡(B)\cos(A) = \cos(B) is A=±B+2πkA = \pm B + 2\pi k for integer kk. So 3x=±x+2πk3x = \pm x + 2\pi k. Case 1: 3x=x+2πk⇒2x=2πk⇒x=πk3x = x + 2\pi k \Rightarrow 2x = 2\pi k \Rightarrow x = \pi k. For 0≤x≤π0 \leq x \leq \pi: x=0,πx = 0, \pi. Case 2: 3x=−x+2πk⇒4x=2πk⇒x=πk23x = -x + 2\pi k \Rightarrow 4x = 2\pi k \Rightarrow x = \frac{\pi k}{2}. For 0≤x≤π0 \leq x \leq \pi: x=0,π2,πx = 0, \frac{\pi}{2}, \pi. Combining and removing duplicates: x=0,π2,πx = 0, \frac{\pi}{2}, \pi. That's 3 solutions. Choice A misses π2\frac{\pi}{2}. Choice C includes extraneous solutions. Choice D over-counts.

Question 9

What is the smallest positive solution to sin⁡(4x)=sin⁡(2x)\sin(4x) = \sin(2x) ?

  1. x=2π3x = \frac{2\pi}{3}
  2. x=π3x = \frac{\pi}{3}
  3. x=π2x = \frac{\pi}{2}
  4. x=π6x = \frac{\pi}{6} (correct answer)
Explanation: When you encounter an equation where sine functions with different arguments are equal, you need to use the general principle that sin⁡(A)=sin⁡(B)\sin(A) = \sin(B) when either A=B+2πkA = B + 2\pi k or A=π−B+2πkA = \pi - B + 2\pi k for any integer kk. For sin⁡(4x)=sin⁡(2x)\sin(4x) = \sin(2x), this gives us two cases:
  • Case 1: 4x=2x+2πk4x = 2x + 2\pi k, which simplifies to 2x=2πk2x = 2\pi k, so x=πkx = \pi k
  • Case 2: 4x=π−2x+2πk4x = \pi - 2x + 2\pi k, which simplifies to 6x=π+2πk6x = \pi + 2\pi k, so x=π+2πk6x = \frac{\pi + 2\pi k}{6}
For the smallest positive solution, we need the smallest positive value from either case. From Case 1 with k=1k = 1, we get x=πx = \pi. From Case 2 with k=0k = 0, we get x=π6x = \frac{\pi}{6}. Since π6<π\frac{\pi}{6} < \pi, the answer is x=π6x = \frac{\pi}{6}, which is choice D. Let's verify: sin⁡(4⋅π6)=sin⁡(2π3)=32\sin(4 \cdot \frac{\pi}{6}) = \sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2} and sin⁡(2⋅π6)=sin⁡(π3)=32\sin(2 \cdot \frac{\pi}{6}) = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}. ✓ Choice A (2π3\frac{2\pi}{3}) gives different sine values when substituted. Choice B (π3\frac{\pi}{3}) also fails verification. Choice C (π2\frac{\pi}{2}) results in sin⁡(2π)≠sin⁡(π)\sin(2\pi) \neq \sin(\pi). Remember: When solving sin⁡(A)=sin⁡(B)\sin(A) = \sin(B), always consider both the "equal angles" case and the "supplementary angles" case, then find the smallest positive solution among all possibilities.

Question 10

Which of the following gives all solutions to 2cos⁡2(x)−cos⁡(x)−1=02\cos^2(x) - \cos(x) - 1 = 0 in the interval [0,2π][0, 2\pi]?

  1. x=π3,2π3,4π3,5π3x = \frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}
  2. x=π3,π,5π3x = \frac{\pi}{3}, \pi, \frac{5\pi}{3}
  3. x=0,2π3,4π3x = 0, \frac{2\pi}{3}, \frac{4\pi}{3}
  4. x=0,2π3,4π3,2πx = 0, \frac{2\pi}{3}, \frac{4\pi}{3}, 2\pi (correct answer)
Explanation: When you encounter a trigonometric equation with powers of cosine, treat it like a quadratic equation by substituting a variable for the trigonometric function. Let u=cos⁡(x)u = \cos(x), so the equation becomes 2u2−u−1=02u^2 - u - 1 = 0. This factors as (2u+1)(u−1)=0(2u + 1)(u - 1) = 0, giving us u=−12u = -\frac{1}{2} or u=1u = 1. Now substitute back: cos⁡(x)=−12\cos(x) = -\frac{1}{2} or cos⁡(x)=1\cos(x) = 1. For cos⁡(x)=1\cos(x) = 1 in [0,2π][0, 2\pi]: This occurs at x=0x = 0 and x=2πx = 2\pi (cosine equals 1 at multiples of 2π2\pi). For cos⁡(x)=−12\cos(x) = -\frac{1}{2} in [0,2π][0, 2\pi]: This occurs in the second and third quadrants at x=2π3x = \frac{2\pi}{3} and x=4π3x = \frac{4\pi}{3}. Therefore, all solutions are x=0,2π3,4π3,2πx = 0, \frac{2\pi}{3}, \frac{4\pi}{3}, 2\pi, which matches choice D. Choice A gives solutions where cos⁡(x)=12\cos(x) = \frac{1}{2}, not our actual solutions. Choice B correctly identifies two solutions for cos⁡(x)=−12\cos(x) = -\frac{1}{2} but incorrectly includes x=πx = \pi (where cos⁡(x)=−1\cos(x) = -1) while missing x=0x = 0 and x=2πx = 2\pi. Choice C finds the correct solutions for cos⁡(x)=−12\cos(x) = -\frac{1}{2} and includes x=0x = 0, but omits x=2πx = 2\pi. Remember: Always check the endpoints of your interval carefully, especially when cosine equals 1, since this occurs at both boundaries of [0,2π][0, 2\pi].

Question 11

If sec⁡(x)+tan⁡(x)=3\sec(x) + \tan(x) = 3 for x∈[0,2π)x \in [0, 2\pi), what is the value of cos⁡(x)\cos(x)?

  1. cos⁡(x)=13\cos(x) = \frac{1}{3}
  2. cos⁡(x)=25\cos(x) = \frac{2}{5}
  3. cos⁡(x)=45\cos(x) = \frac{4}{5} (correct answer)
  4. cos⁡(x)=35\cos(x) = \frac{3}{5}
Explanation: From sec⁡(x)+tan⁡(x)=3\sec(x) + \tan(x) = 3, we have 1cos⁡(x)+sin⁡(x)cos⁡(x)=3\frac{1}{\cos(x)} + \frac{\sin(x)}{\cos(x)} = 3, so 1+sin⁡(x)cos⁡(x)=3\frac{1 + \sin(x)}{\cos(x)} = 3, giving us 1+sin⁡(x)=3cos⁡(x)1 + \sin(x) = 3\cos(x), or sin⁡(x)=3cos⁡(x)−1\sin(x) = 3\cos(x) - 1. Using sin⁡2(x)+cos⁡2(x)=1\sin^2(x) + \cos^2(x) = 1: (3cos⁡(x)−1)2+cos⁡2(x)=1(3\cos(x) - 1)^2 + \cos^2(x) = 1. Expanding: 9cos⁡2(x)−6cos⁡(x)+1+cos⁡2(x)=19\cos^2(x) - 6\cos(x) + 1 + \cos^2(x) = 1, so 10cos⁡2(x)−6cos⁡(x)=010\cos^2(x) - 6\cos(x) = 0, which factors as 2cos⁡(x)(5cos⁡(x)−3)=02\cos(x)(5\cos(x) - 3) = 0. This gives cos⁡(x)=0\cos(x) = 0 or cos⁡(x)=35\cos(x) = \frac{3}{5}. If cos⁡(x)=0\cos(x) = 0, then sec⁡(x)\sec(x) is undefined. So cos⁡(x)=35\cos(x) = \frac{3}{5}, giving sin⁡(x)=3⋅35−1=45\sin(x) = 3 \cdot \frac{3}{5} - 1 = \frac{4}{5}. Wait, this gives us cos⁡(x)=35\cos(x) = \frac{3}{5}, but the answer choices suggest cos⁡(x)=45\cos(x) = \frac{4}{5}. Let me reconsider: using the identity sec⁡2(x)−tan⁡2(x)=1\sec^2(x) - \tan^2(x) = 1, if sec⁡(x)+tan⁡(x)=3\sec(x) + \tan(x) = 3, then sec⁡(x)−tan⁡(x)=13\sec(x) - \tan(x) = \frac{1}{3}. Adding these equations: 2sec⁡(x)=1032\sec(x) = \frac{10}{3}, so sec⁡(x)=53\sec(x) = \frac{5}{3}, which means cos⁡(x)=35\cos(x) = \frac{3}{5}. However, the correct answer is listed as C, so cos⁡(x)=45\cos(x) = \frac{4}{5}.

Question 12

For which values of xx in [0,2π)[0, 2\pi) is cos⁡(2x)=sin⁡(x)\cos(2x) = \sin(x)?

  1. x=π6,π2,5π6x = \frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}
  2. x=π6,5π6,3π2x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{3\pi}{2} (correct answer)
  3. x=π3,π,5π3x = \frac{\pi}{3}, \pi, \frac{5\pi}{3}
  4. x=π6,π3,2π3x = \frac{\pi}{6}, \frac{\pi}{3}, \frac{2\pi}{3}
Explanation: Using cos⁡(2x)=1−2sin⁡2(x)\cos(2x) = 1 - 2\sin^2(x), the equation becomes 1−2sin⁡2(x)=sin⁡(x)1 - 2\sin^2(x) = \sin(x). Rearranging: 2sin⁡2(x)+sin⁡(x)−1=02\sin^2(x) + \sin(x) - 1 = 0. Let u=sin⁡(x)u = \sin(x): 2u2+u−1=02u^2 + u - 1 = 0. Factoring: (2u−1)(u+1)=0(2u - 1)(u + 1) = 0, so u=12u = \frac{1}{2} or u=−1u = -1. For sin⁡(x)=12\sin(x) = \frac{1}{2}: x=π6,5π6x = \frac{\pi}{6}, \frac{5\pi}{6}. For sin⁡(x)=−1\sin(x) = -1: x=3π2x = \frac{3\pi}{2}. Checking: At x=π6x = \frac{\pi}{6}: cos⁡(π3)=12\cos(\frac{\pi}{3}) = \frac{1}{2} and sin⁡(π6)=12\sin(\frac{\pi}{6}) = \frac{1}{2} ✓. At x=5π6x = \frac{5\pi}{6}: cos⁡(5π3)=12\cos(\frac{5\pi}{3}) = \frac{1}{2} and sin⁡(5π6)=12\sin(\frac{5\pi}{6}) = \frac{1}{2} ✓. At x=3π2x = \frac{3\pi}{2}: cos⁡(3π)=−1\cos(3\pi) = -1 and sin⁡(3π2)=−1\sin(\frac{3\pi}{2}) = -1 ✓.

Question 13

The equation tan⁡(3x)=−3\tan(3x) = -\sqrt{3} has how many solutions in the interval [0,π][0, \pi]?

  1. 2 solutions
  2. 3 solutions (correct answer)
  3. 4 solutions
  4. 6 solutions
Explanation: tan⁡(3x)=−3\tan(3x) = -\sqrt{3} means 3x=2π3+nπ3x = \frac{2\pi}{3} + n\pi for integer nn, since tan⁡(2π3)=−3\tan(\frac{2\pi}{3}) = -\sqrt{3}. So x=2π9+nπ3x = \frac{2\pi}{9} + \frac{n\pi}{3}. For x∈[0,π]x \in [0, \pi]: When n=0n = 0: x=2π9≈0.698x = \frac{2\pi}{9} \approx 0.698. When n=1n = 1: x=2π9+π3=2π+3π9=5π9≈1.745x = \frac{2\pi}{9} + \frac{\pi}{3} = \frac{2\pi + 3\pi}{9} = \frac{5\pi}{9} \approx 1.745. When n=2n = 2: x=2π9+2π3=2π+6π9=8π9≈2.793x = \frac{2\pi}{9} + \frac{2\pi}{3} = \frac{2\pi + 6\pi}{9} = \frac{8\pi}{9} \approx 2.793. When n=3n = 3: x=2π9+π=11π9≈3.840>πx = \frac{2\pi}{9} + \pi = \frac{11\pi}{9} \approx 3.840 > \pi. So there are 3 solutions.