Algebra 2 · Grades 10, 11

Solving Trigonometric Equations

Quick answer

A trigonometric equation usually has infinitely many solutions, because the functions repeat. Isolate the trig function, find one angle with the unit circle or an inverse function, use symmetry to find every solution in one period, and add multiples of the period to get them all. Equations that are quadratic in sin x or cos x factor like any quadratic. With 2x or 3x inside, solve for the whole angle over a longer interval, then divide.

What you'll learn

  • Solve basic sine, cosine and tangent equations on one period
  • Write the general solution with multiples of the period
  • Solve trigonometric equations that factor like quadratics
  • Solve equations with multiple angles such as sin 2x

One equation, many solutions

The equation 2sin⁡x=12\sin x = 1 says sin⁡x=12\sin x = \tfrac{1}{2}. On the unit circle, two angles in one turn have a yy-coordinate of 12\tfrac{1}{2}: π6\tfrac{\pi}{6} in the first quadrant and 5π6\tfrac{5\pi}{6} in the second. And because sine repeats every 2π2\pi, adding any number of full turns gives more solutions.

Solutions of sin x = 1/2 The sine wave and the dashed line y = 1/2. They cross twice in every cycle; dots mark the crossings, including π/6 and 5π/6, the two between 0 and 2π, and their repeats 2π to the left and right. -5510-11xy π/6 5π/6
  • y = sin x
  • y = 1/2
Solutions of sin x = 1/2

The method

  1. Isolate the trig function: sin⁡x=12\sin x = \tfrac{1}{2}.
  2. Find the reference angle, the acute angle with that value: π6\tfrac{\pi}{6}.
  3. Place it in every quadrant where the function has the right sign. Sine is positive in quadrants I and II, giving π6\tfrac{\pi}{6} and π−π6=5π6\pi - \tfrac{\pi}{6} = \tfrac{5\pi}{6}.
  4. Add the period for the general solution: x=π6+2kπx = \tfrac{\pi}{6} + 2k\pi or x=5π6+2kπx = \tfrac{5\pi}{6} + 2k\pi, for any integer kk.

Why a calculator gives only one answer

The inverse function sin⁡−1\sin^{-1} must return one angle, so it returns the one between −π2-\tfrac{\pi}{2} and π2\tfrac{\pi}{2}. The unit circle holds the rest. The two points with the same height are mirror images across the yy-axis, so the second solution of sin⁡x=c\sin x = c is π−x\pi - x. For cosine, the two points with the same xx-coordinate are mirror images across the xx-axis, so the second solution is 2π−x2\pi - x. The calculator finds one angle, and the symmetry of the circle finds the others.

Worked examples

Common mistakes

Practice problems

  1. Solve 2cos⁡x=1\sqrt{2}\cos x = 1 on [0,2π)[0, 2\pi).

    Answer

    π4\tfrac{\pi}{4} and 7π4\tfrac{7\pi}{4}

    Full solution

    cos⁡x=22\cos x = \tfrac{\sqrt{2}}{2}, positive in quadrants I and IV: π4\tfrac{\pi}{4} and 2π−π42\pi - \tfrac{\pi}{4}.

  2. Solve 2sin⁡x+3=02\sin x + \sqrt{3} = 0 on [0,2π)[0, 2\pi).

    Answer

    4π3\tfrac{4\pi}{3} and 5π3\tfrac{5\pi}{3}

    Full solution

    sin⁡x=−32\sin x = -\tfrac{\sqrt{3}}{2}. The reference angle is π3\tfrac{\pi}{3}, and sine is negative in quadrants III and IV: π+π3\pi + \tfrac{\pi}{3} and 2π−π32\pi - \tfrac{\pi}{3}.

  3. Solve tan⁡x=−1\tan x = -1 on [0,2π)[0, 2\pi).

    Answer

    3π4\tfrac{3\pi}{4} and 7π4\tfrac{7\pi}{4}

    Full solution

    The reference angle is π4\tfrac{\pi}{4}, and tangent is negative in quadrants II and IV.

  4. Solve sin⁡2x=14\sin^2 x = \tfrac{1}{4} on [0,2π)[0, 2\pi).

    Answer

    π6\tfrac{\pi}{6}, 5π6\tfrac{5\pi}{6}, 7π6\tfrac{7\pi}{6} and 11π6\tfrac{11\pi}{6}

    Full solution

    sin⁡x=±12\sin x = \pm\tfrac{1}{2}: the reference angle π6\tfrac{\pi}{6} in all four quadrants.

  5. Solve 2sin⁡2x−sin⁡x−1=02\sin^2 x - \sin x - 1 = 0 on [0,2π)[0, 2\pi).

    Answer

    π2\tfrac{\pi}{2}, 7π6\tfrac{7\pi}{6} and 11π6\tfrac{11\pi}{6}

    Full solution

    (2sin⁡x+1)(sin⁡x−1)=0(2\sin x + 1)(\sin x - 1) = 0. sin⁡x=1\sin x = 1 gives π2\tfrac{\pi}{2}; sin⁡x=−12\sin x = -\tfrac{1}{2} gives 7π6\tfrac{7\pi}{6} and 11π6\tfrac{11\pi}{6}.

  6. Solve sin⁡xcos⁡x=sin⁡x\sin x\cos x = \sin x on [0,2π)[0, 2\pi).

    Answer

    00 and π\pi

    Full solution

    sin⁡x(cos⁡x−1)=0\sin x(\cos x - 1) = 0. sin⁡x=0\sin x = 0 gives 00 and π\pi; cos⁡x=1\cos x = 1 gives 00 again.

  7. Solve cos⁡2x=0\cos 2x = 0 on [0,2π)[0, 2\pi).

    Answer

    π4\tfrac{\pi}{4}, 3π4\tfrac{3\pi}{4}, 5π4\tfrac{5\pi}{4} and 7π4\tfrac{7\pi}{4}

    Full solution

    With u=2xu = 2x on [0,4π)[0, 4\pi): cos⁡u=0\cos u = 0 at π2,3π2,5π2,7π2\tfrac{\pi}{2}, \tfrac{3\pi}{2}, \tfrac{5\pi}{2}, \tfrac{7\pi}{2}. Divide by 22.

  8. Solve 5cos⁡x=25\cos x = 2 on [0,2π)[0, 2\pi), to three decimal places.

    Answer

    About 1.1591.159 and 5.1245.124

    Full solution

    cos⁡−10.4≈1.159\cos^{-1} 0.4 \approx 1.159. The partner is 2π−1.159≈5.1242\pi - 1.159 \approx 5.124.

  9. Find the general solution of cos⁡x=0\cos x = 0.

    Answer

    x=π2+kπx = \tfrac{\pi}{2} + k\pi

    Full solution

    Cosine is 00 at π2\tfrac{\pi}{2} and 3π2\tfrac{3\pi}{2}, which are π\pi apart, so one family with step π\pi covers both.

  10. A student solves sin⁡x=12\sin x = \tfrac{1}{2} on [0,2π)[0, 2\pi) and answers only x=π6x = \tfrac{\pi}{6}. What went wrong?

    Hint

    In which quadrants is sine positive?

    Answer

    The second solution, 5π6\tfrac{5\pi}{6}, is missing.

    Full solution

    Sine is positive in quadrants I and II, so the reference angle π6\tfrac{\pi}{6} appears twice: π6\tfrac{\pi}{6} and π−π6=5π6\pi - \tfrac{\pi}{6} = \tfrac{5\pi}{6}. The graph shows two crossings of y=12y = \tfrac{1}{2} in every cycle.

Frequently asked questions

Why do trig equations have so many solutions?

Sine, cosine and tangent repeat. If x is a solution of sin x = 1/2, so is x + 2π, x + 4π and so on, and the unit circle gives a second family too.

How do I find all solutions in [0, 2π)?

Find the reference angle, then place it in every quadrant where the function has the right sign. For sin x = 1/2, that is quadrants I and II: π/6 and 5π/6.

What is the general solution?

Every solution, written with an integer k: for sin x = 1/2, x = π/6 + 2kπ or x = 5π/6 + 2kπ.

Can I divide both sides by sin x?

Not safely: that loses the solutions where sin x = 0. Move everything to one side and factor instead.

How do I solve sin 2x = 1/2 on [0, 2π)?

Let u = 2x, which runs over [0, 4π). Solve sin u = 1/2 there, finding four angles, then divide each by 2.

What to learn next

Standards alignment

This lesson covers the following Common Core State Standards for Mathematics.

  • CCSS.MATH.CONTENT.HSF.TF.B.7Trigonometric Functions(+) Use inverse functions to solve trigonometric equations that arise in modeling contexts; evaluate the solutions using technology, and interpret them in terms of the context.
  • CCSS.MATH.CONTENT.HSF.TF.A.2Trigonometric FunctionsExplain how the unit circle in the coordinate plane enables the extension of trigonometric functions to all real numbers, interpreted as radian measures of angles traversed counterclockwise around the unit circle.
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