Trigonometric equations

Basic equations, quadratic equations, factoring and sum-to-product methods. 17 problems with step-by-step solutions.

Solve each problem in your notebook first, then open the solution. In all answers \(k, m \in \mathbb{Z}\).

Problem 1 sin x = a

Solve.

a) \(\sin x = \frac{\sqrt2}{2}\) b) \(2\sin 3x - 1 = 0\) c) \(\sin x = -\frac{\sqrt3}{2}\) d) \(2\sin x + \sqrt2 = 0\) e) \(4\sin x = 0\)

Show solution

\(\sin x = a \iff x = \arcsin a + 2k\pi\) or \(x = \pi - \arcsin a + 2k\pi\) (for \(-1 \le a \le 1\)).

a) \(x = \frac{\pi}{4} + 2k\pi\) or \(x = \frac{3\pi}{4} + 2k\pi\).
b) \(\sin 3x = \frac12\): \(3x = \frac{\pi}{6} + 2k\pi\) or \(3x = \frac{5\pi}{6} + 2k\pi\), so \(x = \frac{\pi}{18} + \frac{2k\pi}{3}\) or \(x = \frac{5\pi}{18} + \frac{2k\pi}{3}\).
c) \(x = -\frac{\pi}{3} + 2k\pi\) or \(x = \frac{4\pi}{3} + 2k\pi\).
d) \(\sin x = -\frac{\sqrt2}{2}\): \(x = -\frac{\pi}{4} + 2k\pi\) or \(x = \frac{5\pi}{4} + 2k\pi\).
e) \(\sin x = 0\): \(x = k\pi\).

Problem 2 sin x = a

Solve.

a) \(2\sin\frac{x}{3} - 1 = 0\) b) \(\sin\left(\frac{x}{2} + \frac{\pi}{6}\right) - \frac{\sqrt3}{2} = 0\) c) \(\sqrt2\sin 2x + 1 = 0\)

Show solution

a) \(\sin\frac{x}{3} = \frac12\): \(\frac{x}{3} = \frac{\pi}{6} + 2k\pi\) or \(\frac{5\pi}{6} + 2k\pi\), so \(x = \frac{\pi}{2} + 6k\pi\) or \(x = \frac{5\pi}{2} + 6k\pi\).

b) \(\frac{x}{2} + \frac{\pi}{6} = \frac{\pi}{3} + 2k\pi\) or \(\frac{2\pi}{3} + 2k\pi\), so \(\frac{x}{2} = \frac{\pi}{6} + 2k\pi\) or \(\frac{\pi}{2} + 2k\pi\): \(x = \frac{\pi}{3} + 4k\pi\) or \(x = \pi + 4k\pi\).

c) \(\sin 2x = -\frac{\sqrt2}{2}\): \(2x = -\frac{\pi}{4} + 2k\pi\) or \(\frac{5\pi}{4} + 2k\pi\), so \(x = -\frac{\pi}{8} + k\pi\) or \(x = \frac{5\pi}{8} + k\pi\).

Problem 3 cos x = a

Solve.

a) \(\cos x = \frac{\sqrt2}{2}\) b) \(2\cos x + 1 = 0\) c) \(\cos x = -\frac{\sqrt3}{2}\) d) \(2\cos 4x - \sqrt3 = 0\) e) \(\cos 4x - 3 = 0\)

Show solution

\(\cos x = a \iff x = \pm\arccos a + 2k\pi\) (for \(-1 \le a \le 1\)).

a) \(x = \pm\frac{\pi}{4} + 2k\pi\).
b) \(\cos x = -\frac12\): \(x = \pm\frac{2\pi}{3} + 2k\pi\).
c) \(x = \pm\frac{5\pi}{6} + 2k\pi\).
d) \(\cos 4x = \frac{\sqrt3}{2}\): \(4x = \pm\frac{\pi}{6} + 2k\pi\), so \(x = \pm\frac{\pi}{24} + \frac{k\pi}{2}\).
e) \(\cos 4x = 3 > 1\): no solution.

Problem 4 cos x = a

Solve.

a) \(\cos\left(2x + \frac{\pi}{3}\right) = 0\) b) \(\sin\left(\frac{\pi}{2} - 2x\right) = -1\) c) \(\cos 6x = -\frac{\sqrt2}{2}\) d) \(\cos\left(x - \frac{\pi}{4}\right) = \frac12\) e) \(2\cos\frac{x}{4} - \sqrt2 = 0\)

Show solution

a) \(2x + \frac{\pi}{3} = \frac{\pi}{2} + k\pi\), so \(x = \frac{\pi}{12} + \frac{k\pi}{2}\).

b) \(\sin\left(\frac{\pi}{2} - 2x\right) = \cos 2x = -1\): \(2x = \pi + 2k\pi\), so \(x = \frac{\pi}{2} + k\pi\).

c) \(6x = \pm\frac{3\pi}{4} + 2k\pi\), so \(x = \pm\frac{\pi}{8} + \frac{k\pi}{3}\).

d) \(x - \frac{\pi}{4} = \pm\frac{\pi}{3} + 2k\pi\), so \(x = \frac{7\pi}{12} + 2k\pi\) or \(x = -\frac{\pi}{12} + 2k\pi\).

e) \(\cos\frac{x}{4} = \frac{\sqrt2}{2}\): \(\frac{x}{4} = \pm\frac{\pi}{4} + 2k\pi\), so \(x = \pm\pi + 8k\pi\).

Problem 5 tan x = a and cot x = a

Solve.

a) \(\tan 2x = 1\) b) \(3\tan\left(2x - \frac{\pi}{4}\right) + 3 = 0\) c) \(\tan(3x - 15^\circ) = 1\) (answer in degrees) d) \(\sqrt3\cot 3x = 1\) e) \(\cot\left(-\frac{x}{3}\right) = 1\)

Show solution

\(\tan x = a \iff x = \arctan a + k\pi\); \(\cot x = a \iff x = \operatorname{arccot} a + k\pi\).

a) \(2x = \frac{\pi}{4} + k\pi\), so \(x = \frac{\pi}{8} + \frac{k\pi}{2}\).
b) \(\tan\left(2x - \frac{\pi}{4}\right) = -1\): \(2x - \frac{\pi}{4} = -\frac{\pi}{4} + k\pi\), so \(x = \frac{k\pi}{2}\).
c) \(3x - 15^\circ = 45^\circ + k\cdot 180^\circ\), so \(x = 20^\circ + k\cdot 60^\circ\).
d) \(\cot 3x = \frac{\sqrt3}{3}\): \(3x = \frac{\pi}{3} + k\pi\), so \(x = \frac{\pi}{9} + \frac{k\pi}{3}\).
e) \(\cot\frac{x}{3} = -1\): \(\frac{x}{3} = \frac{3\pi}{4} + k\pi\), so \(x = \frac{9\pi}{4} + 3k\pi\).

Problem 6 tan x = a and cot x = a

Solve.

a) \(2\tan 4x + 2 = 0\) b) \(\tan\left(\frac{x}{3} - \frac{\pi}{6}\right) = 1\) c) \(\cot(-x) = \sqrt3\) d) \(\cot 3x = 1\) e) \(\sqrt3\tan x + 3 = 0\)

Show solution

a) \(\tan 4x = -1\): \(4x = -\frac{\pi}{4} + k\pi\), so \(x = -\frac{\pi}{16} + \frac{k\pi}{4}\).
b) \(\frac{x}{3} - \frac{\pi}{6} = \frac{\pi}{4} + k\pi\), so \(\frac{x}{3} = \frac{5\pi}{12} + k\pi\) and \(x = \frac{5\pi}{4} + 3k\pi\).
c) \(\cot(-x) = -\cot x\), so \(\cot x = -\sqrt3\) and \(x = \frac{5\pi}{6} + k\pi\).
d) \(3x = \frac{\pi}{4} + k\pi\), so \(x = \frac{\pi}{12} + \frac{k\pi}{3}\).
e) \(\tan x = -\sqrt3\), so \(x = -\frac{\pi}{3} + k\pi\).

Problem 7 Reduction first

Solve.

a) \(\sin\left(\frac{\pi}{2} - \frac{x}{2}\right) = \frac{\sqrt2}{2}\) b) \(\cos\left(\frac{\pi}{2} + 2x\right) = \frac12\) c) \(\tan(\pi + 3x) = \sqrt3\) d) \(\cot\left(\frac{\pi}{2} - 3x\right) = 1\)

Show solution

a) \(\cos\frac{x}{2} = \frac{\sqrt2}{2}\): \(\frac{x}{2} = \pm\frac{\pi}{4} + 2k\pi\), so \(x = \pm\frac{\pi}{2} + 4k\pi\).

b) \(-\sin 2x = \frac12\), so \(\sin 2x = -\frac12\): \(2x = -\frac{\pi}{6} + 2k\pi\) or \(\frac{7\pi}{6} + 2k\pi\), so \(x = -\frac{\pi}{12} + k\pi\) or \(x = \frac{7\pi}{12} + k\pi\).

c) \(\tan 3x = \sqrt3\): \(x = \frac{\pi}{9} + \frac{k\pi}{3}\).

d) \(\tan 3x = 1\): \(x = \frac{\pi}{12} + \frac{k\pi}{3}\).

Problem 8 Solutions in an interval

a) Find all solutions of \(\sin x = \frac12\) with \(400^\circ < x < 900^\circ\).

b) Solve \(\dfrac{\cos 3x}{4} = \dfrac12\).

Show solution

a) \(x = 30^\circ + k\cdot 360^\circ\) or \(x = 150^\circ + k\cdot 360^\circ\).

From the first family: \(390^\circ\) is too small, \(750^\circ\) fits, \(1110^\circ\) is too big. From the second: \(510^\circ\) and \(870^\circ\) fit.

Answer: \(510^\circ,\ 750^\circ,\ 870^\circ\).

b) \(\cos 3x = 2\). Since \(-1 \le \cos 3x \le 1\), there is no solution.

Problem 9 Quadratic equations

Solve.

a) \(2\sin^2 x - 3\sin x + 1 = 0\) b) \(\cos^2 x - \cos x = 0\) c) \(\tan^2 x - 2\tan x - 3 = 0\) d) \(2\cos^2 x + \sin x - 1 = 0\)

Show solution

a) Put \(t = \sin x\): \(2t^2 - 3t + 1 = 0\), \(t = 1\) or \(t = \frac12\). So \(x = \frac{\pi}{2} + 2k\pi\), \(x = \frac{\pi}{6} + 2k\pi\) or \(x = \frac{5\pi}{6} + 2k\pi\).

b) \(\cos x(\cos x - 1) = 0\): \(\cos x = 0\) gives \(x = \frac{\pi}{2} + k\pi\); \(\cos x = 1\) gives \(x = 2k\pi\).

c) \((\tan x - 3)(\tan x + 1) = 0\): \(x = \arctan 3 + k\pi\) or \(x = -\frac{\pi}{4} + k\pi\).

d) \(2(1 - \sin^2 x) + \sin x - 1 = 0 \Rightarrow 2\sin^2 x - \sin x - 1 = 0\), so \(\sin x = 1\) or \(\sin x = -\frac12\): \(x = \frac{\pi}{2} + 2k\pi\), \(x = -\frac{\pi}{6} + 2k\pi\) or \(x = \frac{7\pi}{6} + 2k\pi\).

Problem 10 Quadratic equations

Solve.

a) \(\sin^2 x + \cos x + 1 = 0\) b) \(\cot^2 x - 4\cot x + 3 = 0\) c) \(3\cos 2x + 4\cos x + 1 = 0\) d) \(\cos^2 x + \cos x = 2\) e) \(\cos 2x + 2\tan^2 x = 2\)

Show solution

a) \(1 - \cos^2 x + \cos x + 1 = 0 \Rightarrow \cos^2 x - \cos x - 2 = 0\), so \(\cos x = 2\) (impossible) or \(\cos x = -1\): \(x = \pi + 2k\pi\).

b) \((\cot x - 1)(\cot x - 3) = 0\): \(x = \frac{\pi}{4} + k\pi\) or \(x = \operatorname{arccot} 3 + k\pi\).

c) \(3(2\cos^2 x - 1) + 4\cos x + 1 = 0 \Rightarrow 3\cos^2 x + 2\cos x - 1 = 0\), so \(\cos x = \frac13\) or \(\cos x = -1\): \(x = \pm\arccos\frac13 + 2k\pi\) or \(x = \pi + 2k\pi\).

d) \((\cos x - 1)(\cos x + 2) = 0\), so \(\cos x = 1\): \(x = 2k\pi\).

e) Put \(t = \cos^2 x\) (\(t \ne 0\)): \(\cos 2x = 2t - 1\) and \(\tan^2 x = \frac{1 - t}{t}\). Then \(2t - 1 + \frac{2(1 - t)}{t} = 2 \Rightarrow 2t^2 - 5t + 2 = 0\), so \(t = 2\) (impossible) or \(t = \frac12\). \(\cos^2 x = \frac12\) gives \(x = \frac{\pi}{4} + \frac{k\pi}{2}\).

Problem 11 Factoring

Solve.

a) \(\sin 2x - \cos x = 0\) b) \(\sin x\cos x = \sin x\) c) \(\sin 2x = 2\cos^2 x\) d) \(\sin 2x + 2\cos^2 x = 0\) e) \(2\sin^2 x = \sqrt3\sin x\)

Show solution

a) \(\cos x(2\sin x - 1) = 0\): \(x = \frac{\pi}{2} + k\pi\), \(x = \frac{\pi}{6} + 2k\pi\) or \(x = \frac{5\pi}{6} + 2k\pi\).

b) \(\sin x(\cos x - 1) = 0\): \(\sin x = 0\) gives \(x = k\pi\); \(\cos x = 1\) gives \(x = 2k\pi\), which is already included. Answer \(x = k\pi\).

c) \(2\cos x(\sin x - \cos x) = 0\): \(x = \frac{\pi}{2} + k\pi\) or \(\tan x = 1\), \(x = \frac{\pi}{4} + k\pi\).

d) \(2\cos x(\sin x + \cos x) = 0\): \(x = \frac{\pi}{2} + k\pi\) or \(\tan x = -1\), \(x = -\frac{\pi}{4} + k\pi\).

e) \(\sin x(2\sin x - \sqrt3) = 0\): \(x = k\pi\), \(x = \frac{\pi}{3} + 2k\pi\) or \(x = \frac{2\pi}{3} + 2k\pi\).

Problem 12 Squares

Solve and write the answer in the shortest form.

a) \(4\cos^2 x - 3 = 0\) b) \(\tan^2 x - 3 = 0\) c) \(2\sin^2 x - 1 = 0\) d) \(2\cos^2 x - \cos x - 1 = 0\)

Show solution

a) \(\cos x = \pm\frac{\sqrt3}{2}\): the points \(\pm\frac{\pi}{6}\) and \(\pi \pm \frac{\pi}{6}\) are opposite each other, so \(x = \pm\frac{\pi}{6} + k\pi\).

b) \(\tan x = \pm\sqrt3\): \(x = \pm\frac{\pi}{3} + k\pi\).

c) \(\sin x = \pm\frac{\sqrt2}{2}\): four points \(\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\), equally spaced by \(\frac{\pi}{2}\): \(x = \frac{\pi}{4} + \frac{k\pi}{2}\).

d) \(\cos x = 1\) or \(\cos x = -\frac12\): the points \(0, \frac{2\pi}{3}, \frac{4\pi}{3}\) are equally spaced by \(\frac{2\pi}{3}\): \(x = \frac{2k\pi}{3}\).

Problem 13 Sum-to-product in equations

Solve.

a) \(\cos 5x + \cos 3x = 0\) b) \(\sin 3x = \sin x\) c) \(\cos 4x - \cos 2x = 0\) d) \(\sin x + \sin 3x + \sin 5x = 0\) e) \(\cos x + \cos 2x + \cos 3x = 0\)

Show solution

a) \(2\cos 4x\cos x = 0\): \(x = \frac{\pi}{8} + \frac{k\pi}{4}\) or \(x = \frac{\pi}{2} + k\pi\).

b) \(\sin 3x - \sin x = 2\cos 2x\sin x = 0\): \(x = k\pi\) or \(x = \frac{\pi}{4} + \frac{k\pi}{2}\).

c) \(-2\sin 3x\sin x = 0\): \(x = \frac{k\pi}{3}\) (this includes \(x = k\pi\)).

d) \(\sin x + \sin 5x = 2\sin 3x\cos 2x\), so \(\sin 3x(2\cos 2x + 1) = 0\). \(\sin 3x = 0\) gives \(x = \frac{k\pi}{3}\). \(\cos 2x = -\frac12\) gives \(x = \pm\frac{\pi}{3} + k\pi\), already included. Answer \(x = \frac{k\pi}{3}\).

e) \(\cos x + \cos 3x = 2\cos 2x\cos x\), so \(\cos 2x(2\cos x + 1) = 0\): \(x = \frac{\pi}{4} + \frac{k\pi}{2}\) or \(x = \pm\frac{2\pi}{3} + 2k\pi\).

Problem 14 Equal functions

Solve.

a) \(\sin 3x = \cos x\) b) \(\sin\left(2x + \frac{\pi}{6}\right) = \sin\left(x - \frac{\pi}{3}\right)\) c) \(\cos 2x = \cos 50^\circ\) d) \(\sin 3x = \sin 45^\circ\) e) \(\tan\left(3x + \frac{\pi}{4}\right) = \cot\left(x - \frac{\pi}{6}\right)\)

Show solution

\(\sin A = \sin B \iff A = B + 2k\pi\) or \(A = \pi - B + 2k\pi\); \(\cos A = \cos B \iff A = \pm B + 2k\pi\); \(\tan A = \tan B \iff A = B + k\pi\).

a) \(\sin 3x = \sin\left(\frac{\pi}{2} - x\right)\): \(3x = \frac{\pi}{2} - x + 2k\pi \Rightarrow x = \frac{\pi}{8} + \frac{k\pi}{2}\), or \(3x = \frac{\pi}{2} + x + 2k\pi \Rightarrow x = \frac{\pi}{4} + k\pi\).

b) \(2x + \frac{\pi}{6} = x - \frac{\pi}{3} + 2k\pi \Rightarrow x = -\frac{\pi}{2} + 2k\pi\), or \(2x + \frac{\pi}{6} = \pi - x + \frac{\pi}{3} + 2k\pi \Rightarrow 3x = \frac{7\pi}{6} + 2k\pi \Rightarrow x = \frac{7\pi}{18} + \frac{2k\pi}{3}\).

c) \(2x = \pm 50^\circ + k\cdot 360^\circ\), so \(x = \pm 25^\circ + k\cdot 180^\circ\).

d) \(3x = 45^\circ + k\cdot 360^\circ\) or \(3x = 135^\circ + k\cdot 360^\circ\), so \(x = 15^\circ + k\cdot 120^\circ\) or \(x = 45^\circ + k\cdot 120^\circ\).

e) \(\cot\left(x - \frac{\pi}{6}\right) = \tan\left(\frac{\pi}{2} - x + \frac{\pi}{6}\right) = \tan\left(\frac{2\pi}{3} - x\right)\). So \(3x + \frac{\pi}{4} = \frac{2\pi}{3} - x + k\pi\), \(4x = \frac{5\pi}{12} + k\pi\), \(x = \frac{5\pi}{48} + \frac{k\pi}{4}\). For these \(x\) both sides are defined.

Problem 15 More equations

Solve.

a) \(\tan 3x + \tan x = 0\) b) \(\sin(x + 30^\circ) - \sin(x - 30^\circ) = 0\) c) \(\sin^2 4x + \sin^2 3x = \sin^2 2x + \sin^2 x\)

Show solution

a) \(\tan 3x = -\tan x = \tan(-x)\), so \(3x = -x + k\pi\) and \(x = \frac{k\pi}{4}\). Both tangents must exist, so \(\cos x \ne 0\) and \(\cos 3x \ne 0\): this removes \(x = \frac{\pi}{2} + m\pi\) (\(k = 2, 6, 10, \dots\)). Answer: \(x = m\pi\) or \(x = \frac{\pi}{4} + \frac{m\pi}{2}\).

b) \(2\cos x\sin 30^\circ = 0\), so \(\cos x = 0\): \(x = 90^\circ + k\cdot 180^\circ\).

c) Use \(\sin^2 a = \frac{1 - \cos 2a}{2}\): \(\cos 8x + \cos 6x = \cos 4x + \cos 2x\), so \(2\cos 7x\cos x = 2\cos 3x\cos x\), i.e. \(\cos x(\cos 7x - \cos 3x) = 0\), i.e. \(-2\cos x\sin 5x\sin 2x = 0\).

\(\cos x = 0\): \(x = \frac{\pi}{2} + k\pi\); \(\sin 2x = 0\): \(x = \frac{k\pi}{2}\) (contains the first); \(\sin 5x = 0\): \(x = \frac{k\pi}{5}\). Answer: \(x = \frac{k\pi}{2}\) or \(x = \frac{k\pi}{5}\).

Problem 16 Sum of the roots

a) Find the sum of the roots of \(\sin^4 x + \cos^4 x = \frac58\) in \([0, \pi]\).

b) Find the sum of the roots of \(\dfrac{1}{\sin^2 2x} + \dfrac{1}{\cos^2 2x} = \dfrac{16}{3}\) in \(\left[0, \frac{\pi}{2}\right]\).

c) Find the sum of the roots of \(2\sin\frac{x}{2}\sin\frac{3x}{2} - \sin\frac{3x}{2} = 0\) in \([0, 2\pi]\).

Show solution

a) \(\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x\cos^2 x = 1 - \frac12\sin^2 2x\). So \(\sin^2 2x = \frac34\), \(\sin 2x = \pm\frac{\sqrt3}{2}\), \(x = \pm\frac{\pi}{6} + \frac{k\pi}{2}\).

In \([0, \pi]\): \(\frac{\pi}{6}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{5\pi}{6}\). Sum: \(\frac{(1 + 2 + 4 + 5)\pi}{6} = 2\pi\).

b) \(\frac{\cos^2 2x + \sin^2 2x}{\sin^2 2x\cos^2 2x} = \frac{4}{\sin^2 4x} = \frac{16}{3}\), so \(\sin^2 4x = \frac34\) and \(x = \pm\frac{\pi}{12} + \frac{k\pi}{4}\).

In \(\left[0, \frac{\pi}{2}\right]\): \(\frac{\pi}{12}, \frac{\pi}{6}, \frac{\pi}{3}, \frac{5\pi}{12}\) (none of them makes \(\sin 2x\) or \(\cos 2x\) zero). Sum: \(\frac{(1 + 2 + 4 + 5)\pi}{12} = \pi\).

c) \(\sin\frac{3x}{2}\left(2\sin\frac{x}{2} - 1\right) = 0\). \(\sin\frac{3x}{2} = 0\): \(x = \frac{2k\pi}{3}\), in the interval \(0, \frac{2\pi}{3}, \frac{4\pi}{3}, 2\pi\). \(\sin\frac{x}{2} = \frac12\) with \(\frac{x}{2} \in [0, \pi]\): \(x = \frac{\pi}{3}\) or \(\frac{5\pi}{3}\).

Sum: \(\frac{(0 + 2 + 4 + 6 + 1 + 5)\pi}{3} = 6\pi\).

Problem 17 Special equations

Solve.

a) \(9^{\,2\cos x - 1} = 1\) b) \(\sin^2 3x + \sin^2 x = 1\) c) \(\sin^4 x + \cos^4 x = \cos^2 2x\) d) \(1 + \cos x = 2\sin^2\frac{x}{2}\)

Show solution

a) \(2\cos x - 1 = 0\), \(\cos x = \frac12\): \(x = \pm\frac{\pi}{3} + 2k\pi\).

b) \(\frac{1 - \cos 6x}{2} + \frac{1 - \cos 2x}{2} = 1 \Rightarrow \cos 6x + \cos 2x = 0 \Rightarrow 2\cos 4x\cos 2x = 0\): \(x = \frac{\pi}{8} + \frac{k\pi}{4}\) or \(x = \frac{\pi}{4} + \frac{k\pi}{2}\).

c) \(1 - \frac12\sin^2 2x = 1 - \sin^2 2x\), so \(\sin^2 2x = 0\): \(x = \frac{k\pi}{2}\).

d) \(1 + \cos x = 2\cos^2\frac{x}{2}\), so \(\cos^2\frac{x}{2} = \sin^2\frac{x}{2}\), i.e. \(\cos x = 0\): \(x = \frac{\pi}{2} + k\pi\).