Teach Me Maths

Solving Trigonometric Equations

Solve single-angle trigonometric equations over a stated interval, including quadratic equations and equations using identities, without losing solutions.

Solve for the ratio, then find every angle

First rearrange or factorise to obtain values of sine, cosine or tangent. Then use exact values or an inverse trig function to find a reference angle. Use the graph or quadrant signs to find every angle in the stated interval. Work in degrees.

Sine and cosine must lie between −1 and 1. Reject an impossible ratio before finding angles. Check whether the interval includes its endpoints: 0° and 360° have the same ratios but are different values of the unknown.

See solutions as graph intersections

For \sin x=k on 0^\circ\le x\le360^\circ, the number of solutions changes at k=0 and at the extreme values k=\pm1. Listing angles is safer than assuming there are always two.

Worked example 1

Solve \sin x=1/2 for 0^\circ\le x\le360^\circ.

  1. The reference angle is 30^\circ. Sine is positive in quadrants I and II.

  2. Your turn. The first solution is 30^\circ. Find the second, 180^\circ-30^\circ, in degrees.

    The second solution is 150^\circ

  3. Answer: x=30^\circ,150^\circ. Neither endpoint has sine 1/2

Worked example 2

Solve \cos x=-\sqrt2/2 for 0^\circ\le x\le360^\circ.

  1. The reference angle is 45^\circ. Cosine is negative in quadrants II and III, so use 180^\circ-45^\circ and 180^\circ+45^\circ

  2. Your turn. Find the quadrant III solution, in degrees.

    The quadrant III solution is 225^\circ

  3. Answer: x=135^\circ,225^\circ

Worked example 3

Solve \tan x=-\sqrt3 for 0^\circ\le x\le360^\circ.

  1. The reference angle is 60^\circ. Tangent is negative in quadrants II and IV, giving 180^\circ-60^\circ and 360^\circ-60^\circ

  2. Your turn. Find the quadrant IV solution, in degrees.

    The quadrant IV solution is 300^\circ

  3. Answer: x=120^\circ,300^\circ. The solutions are 180^\circ apart, matching tangent’s period.

Worked example 4

Solve \sin x=-0.4 for 0^\circ\le x\le360^\circ, giving angles to 1 decimal place.

  1. The acute reference angle is \alpha=\sin^{-1}(0.4)=23.5781\ldots^\circ. Sine is negative in quadrants III and IV.

  2. Your turn. Calculate 180^\circ+\alpha to 1 decimal place.

    The quadrant III solution is 203.6^\circ (1 d.p.).

  3. The quadrant IV solution is 360^\circ-\alpha=336.4218\ldots^\circ. A calculator may return -23.5781\ldots^\circ for \sin^{-1}(-0.4); that is outside the requested interval.
  4. Answer: x=203.6^\circ,336.4^\circ (1 d.p.).

Treat the trig ratio as one unknown

An equation such as 2\cos^2x-\cos x-1=0 is quadratic in \cos x. Substitute u=\cos x, solve for u, then find angles for each allowed value. If the equation contains both squared sine and squared cosine, use \sin^2x+\cos^2x=1 to eliminate one.

Factorise instead of dividing by a trig expression that might be zero. Otherwise you can lose an entire set of solutions.

Worked example 5

Solve 2\cos^2x-\cos x-1=0 for 0^\circ\le x\le360^\circ.

  1. Let u=\cos x. Then 2u^2-u-1=(2u+1)(u-1)=0

  2. Your turn. One value is u=1. Find the other value from 2u+1=0

    The other value is u=-1/2. Both values are in the allowed cosine range.

  3. \cos x=1 gives x=0^\circ,360^\circ. The equation \cos x=-1/2 gives x=120^\circ,240^\circ
  4. Answer: x=0^\circ,120^\circ,240^\circ,360^\circ. Both endpoints belong to the given interval.

Worked example 6

Solve 2\sin^2x-\sin x=0 for 0^\circ<x<360^\circ.

  1. Factorise: \sin x(2\sin x-1)=0. Thus \sin x=0 or \sin x=1/2. Do not divide by \sin x

  2. Your turn. Which angle strictly between 0^\circ and 360^\circ has sine zero?

    x=180^\circ. The endpoints 0^\circ and 360^\circ are excluded.

  3. \sin x=1/2 gives 30^\circ and 150^\circ
  4. Answer: x=30^\circ,150^\circ,180^\circ. Dividing by sine would have lost 180^\circ

Worked example 7

Solve \sin^2x=3\cos^2x for 0^\circ\le x\le360^\circ.

  1. Replace \sin^2x by 1-\cos^2x. Then 1-\cos^2x=3\cos^2x, so 4\cos^2x=1

  2. Your turn. \cos^2x=1/4. What is the positive value of \cos x?

    The two values are \cos x=1/2 and \cos x=-1/2

  3. The positive value gives 60^\circ,300^\circ; the negative value gives 120^\circ,240^\circ. All four are in the interval.
  4. Answer: x=60^\circ,120^\circ,240^\circ,300^\circ. Taking only the positive square root would lose two solutions.

Common mistakes

  • Stopping at the calculator’s first angle. Use the graph or symmetry to find every solution in the stated interval.
  • Choosing the wrong quadrant. Use the ratio’s sign to identify the allowed quadrants.
  • Accepting an impossible sine or cosine value. Reject values outside [-1,1].
  • Dividing away a possible zero factor. Factorise and solve each factor equal to zero instead.
  • Taking only one square-root sign. Keep both signs, then check which values satisfy the equation.
  • Mishandling interval endpoints. Check every angle against the stated inclusive or exclusive boundaries.

Now try it: Solving Trigonometric Equations practice questions

More on this topic: Solving Trigonometric Equations worksheet with full solutions