Trigonometric Functions • Topic 3 of 3

Trigonometric Equations

A trigonometric equation involves trigonometric functions of an unknown angle. Because the functions are periodic, such equations have infinitely many solutions, so we report two things: the principal solutions (those in $[0, 2\pi)$) and the general solution (a formula covering all of them).

The general solutions of the three basic equations are standard results, where $n$ is any integer ($n \in \mathbb{Z}$):

$$\sin x = \sin y \quad\Rightarrow\quad x = n\pi + (-1)^n y$$
$$\cos x = \cos y \quad\Rightarrow\quad x = 2n\pi \pm y$$
$$\tan x = \tan y \quad\Rightarrow\quad x = n\pi + y$$

The form of each general solution mirrors the symmetry of the curve. The cosine curve is even and symmetric about the $x$-axis, so its solutions come in $\pm$ pairs spaced by full turns. The sine curve repeats with the alternating reflection captured by $(-1)^n$. The tangent function has the shortest period, $\pi$, so its solutions are spaced just $\pi$ apart. Two useful corollaries follow at once: $\sin^2 x = \sin^2 y$, $\cos^2 x = \cos^2 y$ and $\tan^2 x = \tan^2 y$ each give $x = n\pi \pm y$, because squaring removes the sign information that distinguished the three basic forms.

Some equalities are worth memorising directly:

EquationGeneral solutionReason
$\sin x = 0$$x = n\pi$sine vanishes at every multiple of $\pi$
$\cos x = 0$$x = (2n + 1)\dfrac{\pi}{2}$cosine vanishes at odd multiples of $\dfrac{\pi}{2}$
$\tan x = 0$$x = n\pi$same zeros as sine
$\sin x = 1$$x = (4n+1)\dfrac{\pi}{2}$sine peaks once per turn
$\cos x = 1$$x = 2n\pi$cosine peaks at each full turn

A reliable working method: first reduce the equation to the form $\sin x = k$, $\cos x = k$ or $\tan x = k$; find an angle $y$ whose function value is $k$ (the reference angle); fix the correct quadrant using the sign of $k$; then write the general solution from the formula above. For conditional identities — relations true only when angles satisfy a side-condition such as $A + B + C = \pi$ (the angles of a triangle) — replace one angle using the condition (e.g. $C = \pi - (A+B)$, so $\cos C = -\cos(A+B)$) before applying sum/product formulae to fold the expression into the required form.

Deeper Insight — why three different formulas, and where the $(-1)^n$ comes from: The three general-solution forms look unrelated but each is forced by exactly how many times, and where, a horizontal line $y = k$ cuts the graph in one period. A horizontal line crosses the cosine curve at two points placed symmetrically about $0$, which is precisely what $2n\pi \pm y$ encodes. The same line cuts the sine curve at two points too, but they sit symmetrically about $\tfrac{\pi}{2}$, not $0$ — and writing that asymmetry compactly is exactly what the alternating factor $(-1)^n$ achieves, flipping the sign on every other branch. Tangent, having period $\pi$ and crossing each level just once per period, needs only the simple $n\pi + y$. The danger in this topic is mechanical: blindly applying $x = n\pi + (-1)^n y$ to a cosine equation produces wrong answers, and squaring an equation can introduce extraneous roots that must be checked against the original. Always identify which function you actually have before reaching for a formula, and your solutions will be both complete and correct.

A horizontal line cutting the sine curve at many solutions sin x = k has infinitely many solutions y = k each crossing is one solution → x = nπ + (−1)ⁿy Symmetric cosine solutions about the origin cos x = k: two solutions ±y per turn −y+y
Solved Examples
1
Worked Example
Find the principal solutions of $\sin x = \dfrac{1}{2}$.
Solution
  1. $\sin x$ is positive, so $x$ lies in Quadrants I and II.
  2. The reference angle with sine $\dfrac{1}{2}$ is $\dfrac{\pi}{6}$ (Q1 solution).
  3. The Q2 solution is $\pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$.

Answer: $x = \dfrac{\pi}{6}$ and $x = \dfrac{5\pi}{6}$.

2
Worked Example
Find the principal solutions of $\tan x = \sqrt{3}$.
Solution
  1. $\tan x$ is positive, so $x$ lies in Quadrants I and III.
  2. The reference angle is $\dfrac{\pi}{3}$ (since $\tan\dfrac{\pi}{3} = \sqrt{3}$).
  3. The Q3 solution is $\pi + \dfrac{\pi}{3} = \dfrac{4\pi}{3}$.

Answer: $x = \dfrac{\pi}{3}$ and $x = \dfrac{4\pi}{3}$.

3
Worked Example
Find the general solution of $\cos x = -\dfrac{1}{2}$.
Solution
  1. Find an angle whose cosine is $-\dfrac{1}{2}$: that is $\dfrac{2\pi}{3}$ (cosine negative, Q2).
  2. So the equation is $\cos x = \cos\dfrac{2\pi}{3}$.
  3. Apply $\cos x = \cos y \Rightarrow x = 2n\pi \pm y$.

Answer: $x = 2n\pi \pm \dfrac{2\pi}{3},\ n \in \mathbb{Z}$.

4
Worked Example
Find the general solution of $\sin x = -\dfrac{\sqrt{3}}{2}$.
Solution
  1. An angle with this sine is $-\dfrac{\pi}{3}$, since $\sin\left(-\dfrac{\pi}{3}\right) = -\dfrac{\sqrt{3}}{2}$.
  2. So $\sin x = \sin\left(-\dfrac{\pi}{3}\right)$.
  3. Apply $\sin x = \sin y \Rightarrow x = n\pi + (-1)^n y$.

Answer: $x = n\pi + (-1)^n\left(-\dfrac{\pi}{3}\right) = n\pi - (-1)^n\dfrac{\pi}{3},\ n \in \mathbb{Z}$.

5
Worked Example
Solve $2\cos^2 x + 3\sin x = 0$ (general solution).
Solution
  1. Replace $\cos^2 x = 1 - \sin^2 x$: $2(1 - \sin^2 x) + 3\sin x = 0$.
  2. $-2\sin^2 x + 3\sin x + 2 = 0$, i.e. $2\sin^2 x - 3\sin x - 2 = 0$.
  3. Factorise: $(2\sin x + 1)(\sin x - 2) = 0$, so $\sin x = -\dfrac{1}{2}$ or $\sin x = 2$.
  4. $\sin x = 2$ is impossible ($|\sin x| \le 1$), so take $\sin x = -\dfrac{1}{2} = \sin\left(-\dfrac{\pi}{6}\right)$.
  5. $x = n\pi + (-1)^n\left(-\dfrac{\pi}{6}\right)$.

Answer: $x = n\pi - (-1)^n\dfrac{\pi}{6},\ n \in \mathbb{Z}$.

6
Worked Example
Solve $\tan 2x = -\cot\left(x + \dfrac{\pi}{3}\right)$ (general solution).
Solution
  1. Write $-\cot\theta = \tan\left(\dfrac{\pi}{2} + \theta\right)$, so the right side is $\tan\left(\dfrac{\pi}{2} + x + \dfrac{\pi}{3}\right) = \tan\left(x + \dfrac{5\pi}{6}\right)$.
  2. The equation becomes $\tan 2x = \tan\left(x + \dfrac{5\pi}{6}\right)$.
  3. Apply $\tan x = \tan y \Rightarrow$ angles differ by $n\pi$: $2x = n\pi + x + \dfrac{5\pi}{6}$.
  4. $x = n\pi + \dfrac{5\pi}{6}$.

Answer: $x = n\pi + \dfrac{5\pi}{6},\ n \in \mathbb{Z}$.

7
Worked Example
Find the general solution of $\sin 2x = \dfrac{\sqrt3}{2}$.
Solution
  1. Note $\dfrac{\sqrt3}{2} = \sin\dfrac{\pi}{3}$, so $\sin 2x = \sin\dfrac{\pi}{3}$.
  2. Apply $\sin\theta = \sin y \Rightarrow \theta = n\pi + (-1)^n y$ with $\theta = 2x$, $y = \dfrac{\pi}{3}$.
  3. $2x = n\pi + (-1)^n\dfrac{\pi}{3}$.
  4. Divide by $2$: $x = \dfrac{n\pi}{2} + (-1)^n\dfrac{\pi}{6}$.

Answer: $x = \dfrac{n\pi}{2} + (-1)^n\dfrac{\pi}{6},\ n \in \mathbb{Z}$.

8
Worked Example
Solve $2\sin^2 x = 3\cos x$ for the general solution.
Solution
  1. Replace $\sin^2 x = 1 - \cos^2 x$: $2(1 - \cos^2 x) = 3\cos x$.
  2. $2\cos^2 x + 3\cos x - 2 = 0$.
  3. Factorise: $(2\cos x - 1)(\cos x + 2) = 0$, so $\cos x = \dfrac{1}{2}$ or $\cos x = -2$.
  4. $\cos x = -2$ is impossible, so $\cos x = \dfrac{1}{2} = \cos\dfrac{\pi}{3}$.
  5. $x = 2n\pi \pm \dfrac{\pi}{3}$.

Answer: $x = 2n\pi \pm \dfrac{\pi}{3},\ n \in \mathbb{Z}$.

9
Worked Example
Find the general solution of $\sin x + \sin 3x + \sin 5x = 0$.
Solution
  1. Group the outer terms and use sum-to-product: $\sin x + \sin 5x = 2\sin 3x\cos 2x$.
  2. The equation becomes $2\sin 3x\cos 2x + \sin 3x = 0$, i.e. $\sin 3x(2\cos 2x + 1) = 0$.
  3. Either $\sin 3x = 0 \Rightarrow 3x = n\pi \Rightarrow x = \dfrac{n\pi}{3}$.
  4. Or $\cos 2x = -\dfrac{1}{2} = \cos\dfrac{2\pi}{3} \Rightarrow 2x = 2n\pi \pm \dfrac{2\pi}{3} \Rightarrow x = n\pi \pm \dfrac{\pi}{3}$.

Answer: $x = \dfrac{n\pi}{3}$ or $x = n\pi \pm \dfrac{\pi}{3},\ n \in \mathbb{Z}$.

10
Worked Example
In any triangle $ABC$ (so $A + B + C = \pi$), prove that $\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C$.
Solution
  1. Combine the first two with sum-to-product: $\sin 2A + \sin 2B = 2\sin(A+B)\cos(A-B)$.
  2. Since $A + B = \pi - C$, $\sin(A+B) = \sin C$; so this is $2\sin C\cos(A-B)$.
  3. Also $\sin 2C = 2\sin C\cos C$, and $\cos C = -\cos(A+B)$. The left side becomes $2\sin C\big[\cos(A-B) - \cos(A+B)\big]$.
  4. Use $\cos(A-B) - \cos(A+B) = 2\sin A\sin B$.
  5. Thus the sum $= 2\sin C\cdot 2\sin A\sin B = 4\sin A\sin B\sin C$.

Answer: Proved: $\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C$.

11
Worked Example
Solve $\cos x + \cos 3x = 0$ for the general solution.
Solution
  1. Apply sum-to-product: $\cos x + \cos 3x = 2\cos 2x\cos x$.
  2. So $2\cos 2x\cos x = 0$, giving $\cos 2x = 0$ or $\cos x = 0$.
  3. $\cos 2x = 0 \Rightarrow 2x = (2n+1)\dfrac{\pi}{2} \Rightarrow x = (2n+1)\dfrac{\pi}{4}$.
  4. $\cos x = 0 \Rightarrow x = (2n+1)\dfrac{\pi}{2}$, which is already included in the first family.

Answer: $x = (2n+1)\dfrac{\pi}{4},\ n \in \mathbb{Z}$.

12
Worked Example
If $A + B + C = \pi$, prove the conditional identity $\tan A + \tan B + \tan C = \tan A\tan B\tan C$.
Solution
  1. From $A + B + C = \pi$, $A + B = \pi - C$, so $\tan(A+B) = \tan(\pi - C) = -\tan C$.
  2. Expand the left side: $\dfrac{\tan A + \tan B}{1 - \tan A\tan B} = -\tan C$.
  3. Cross-multiply: $\tan A + \tan B = -\tan C(1 - \tan A\tan B) = -\tan C + \tan A\tan B\tan C$.
  4. Move $\tan C$ across: $\tan A + \tan B + \tan C = \tan A\tan B\tan C$.

Answer: Proved: $\tan A + \tan B + \tan C = \tan A\tan B\tan C$ when $A+B+C = \pi$.

Key Points

  • Periodicity means trig equations have infinitely many solutions; report principal solutions in $[0, 2\pi)$ plus a general solution.
  • $\sin x = \sin y \Rightarrow x = n\pi + (-1)^n y$; the $(-1)^n$ encodes the sine curve's alternating symmetry.
  • $\cos x = \cos y \Rightarrow x = 2n\pi \pm y$; the $\pm$ reflects cosine's symmetry about the origin.
  • $\tan x = \tan y \Rightarrow x = n\pi + y$, the simplest form because $\tan$ has period $\pi$.
  • Squared equations $\sin^2 x = \sin^2 y$ (and likewise for $\cos,\tan$) give $x = n\pi \pm y$.
  • Always identify the function first, reduce to $\sin/\cos/\tan x = k$, then apply the matching formula; discard roots with $|\sin|,|\cos| > 1$.
  • Sum-to-product factoring turns $\sin x + \sin 3x + \sin 5x = 0$ type equations into a product set equal to zero.
  • For conditional identities use the side-condition (e.g. $A+B+C = \pi \Rightarrow \tan(A+B) = -\tan C$) before applying sum/product formulae.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The general solution of $\sin\theta=0$ is:
Explanation: $\sin\theta=0\Rightarrow\theta=n\pi$.
Q2.The general solution of $\cos\theta=0$ is:
Explanation: $\cos\theta=0\Rightarrow\theta=(2n+1)\tfrac{\pi}{2}$.
Q3.If $\sin\theta=\sin\alpha$, then $\theta=$
Explanation: Standard sine general solution.
Q4.The general solution of $\tan\theta=0$ is:
Explanation: $\tan\theta=0\Rightarrow\theta=n\pi$.
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