Solve trigonometric equations in multiples of the angle

Solve trigonometric equations in multiples of the angle

Solving trigonometric equations in multiples of the angle involves substituting u=nθu = n\theta to find all solutions within a transformed interval, then dividing by nn to recover the values of θ\theta.

Equations of the form sin⁡nθ=k\sin n\theta = k, cos⁡nθ=k\cos n\theta = k and tan⁡nθ=p\tan n\theta = p are called trigonometric equations in multiples of the angle. The unknown appears as a multiple nθn\theta rather than as θ\theta itself, and nn is usually a small positive integer such as 22 or 33.

The substitution method

The standard approach is to replace the multiple with a single letter:

  • Let u=nθu = n\theta.
  • Transform the given interval for θ\theta into an interval for uu by multiplying both endpoints by nn.
  • Solve the simpler equation for uu over that new interval.
  • Convert each solution back using θ=un\theta = \dfrac{u}{n}.

Finding all solutions in the interval

A trigonometric equation normally has infinitely many solutions, so the interval matters. Use the symmetry and periodicity of each function:

  • For sin⁡u=k\sin u = k, one solution is u=sin⁡−1ku = \sin^{-1} k; the second in a full turn is u=180∘−sin⁡−1ku = 180^\circ - \sin^{-1} k.
  • For cos⁡u=k\cos u = k, the two solutions in a full turn are u=cos⁡−1ku = \cos^{-1} k and u=360∘−cos⁡−1ku = 360^\circ - \cos^{-1} k.
  • For tan⁡u=p\tan u = p, the tangent repeats every 180∘180^\circ, so add multiples of 180∘180^\circ to u=tan⁡−1pu = \tan^{-1} p.

Because the interval for uu is nn times as wide as the interval for θ\theta, it typically contains nn times as many solutions — for example, an interval of length 360∘360^\circ for u=3θu = 3\theta becomes a range of 1080∘1080^\circ, so expect several solutions before dividing back.

Equations mixing two multiples

When an equation contains two different multiples, such as 2sin⁡2θ=cos⁡2θ2\sin 2\theta = \cos 2\theta, first reduce it to a single trigonometric ratio. Dividing both sides by cos⁡2θ\cos 2\theta gives 2tan⁡2θ=12\tan 2\theta = 1, which is then solved by the substitution method above. Always check that dividing by a cosine does not discard valid solutions where that cosine is zero.

Checking the answer

After dividing each uu-value by nn, confirm that every resulting θ\theta lies inside the original interval. Discard any that fall outside, and state the final answers to the required degree of accuracy.