Trigonometric Equations and General Solutions | NIMCET

Quick answer: The principal solutions of a trigonometric equation lie in $[0, 2\pi)$ ; the general solution lists every solution using an integer $n$ . The three basic results are: $\sin x = \sin\alpha \Rightarrow x = n\pi + (-1)^n\alpha$ ; $\cos x = \cos\alpha \Rightarrow x = 2n\pi \pm \alpha$ ; $\tan x = \tan\alpha \Rightarrow x = n\pi + \alpha$ , where $n \in \mathbb{Z}$ .

Since trigonometric functions are periodic, an equation like $\sin x = \frac{1}{2}$ has infinitely many solutions. The general solution packs all of them into one formula.

Basic general solutions

Equation	General solution ( $n \in \mathbb{Z}$ )
$\sin x = 0$	$x = n\pi$
$\cos x = 0$	$x = (2n + 1)\frac{\pi}{2}$
$\tan x = 0$	$x = n\pi$
$\sin x = \sin\alpha$	$x = n\pi + (-1)^n\alpha$
$\cos x = \cos\alpha$	$x = 2n\pi \pm \alpha$
$\tan x = \tan\alpha$	$x = n\pi + \alpha$
$\sin^2 x = \sin^2\alpha$ , $\cos^2 x = \cos^2\alpha$ or $\tan^2 x = \tan^2\alpha$	$x = n\pi \pm \alpha$
$\sin x = 1$	$x = 2n\pi + \frac{\pi}{2}$
$\cos x = -1$	$x = (2n + 1)\pi$

Why sin has $(-1)^n$ : $\sin x = \sin\alpha$ is satisfied by $\alpha$ and $\pi - \alpha$ , then every $2\pi$ after that. Writing $n$ even gives $2k\pi + \alpha$ and $n$ odd gives $(2k + 1)\pi - \alpha$ , which is exactly $n\pi + (-1)^n\alpha$ .

Types of equations and how to solve them

1. Equations in one function

Rewrite everything in one ratio, solve as an algebraic equation, then reject impossible values.

Example: $2\cos^2 x + 3\sin x = 0$ .

Step 1: $2(1 - \sin^2 x) + 3\sin x = 0$ , so $2\sin^2 x - 3\sin x - 2 = 0$ .

Step 2: $(2\sin x + 1)(\sin x - 2) = 0$ . Reject $\sin x = 2$ .

Step 3: $\sin x = -\frac{1}{2} = \sin\left(-\frac{\pi}{6}\right)$ , so

$x = n\pi + (-1)^n\left(-\frac{\pi}{6}\right) = n\pi - (-1)^n\frac{\pi}{6}$

2. Factorisation using sum-to-product

Example: $\cos x + \cos 3x = 0$ becomes $2\cos 2x\cos x = 0$ , so $\cos 2x = 0$ or $\cos x = 0$ . This gives $x = \frac{(2n + 1)\pi}{4}$ or $x = \frac{(2n + 1)\pi}{2}$ . The sum-to-product formulas are the main tool here.

3. Equations of the form a cos x + b sin x = c

Divide by $r = \sqrt{a^2 + b^2}$ and write the left side as $\cos(x - \alpha)$ , where $\cos\alpha = \frac{a}{r}$ and $\sin\alpha = \frac{b}{r}$ :

$\cos(x - \alpha) = \frac{c}{r}$

A solution exists only if $\lvert c\rvert \le \sqrt{a^2 + b^2}$ .

Example: $\sqrt{3}\cos x + \sin x = \sqrt{2}$ .

Step 1: $r = 2$ , so $\frac{\sqrt{3}}{2}\cos x + \frac{1}{2}\sin x = \frac{1}{\sqrt{2}}$ .

Step 2: The left side is $\cos\left(x - \frac{\pi}{6}\right)$ , and $\frac{1}{\sqrt{2}} = \cos\frac{\pi}{4}$ .

Step 3: $x - \frac{\pi}{6} = 2n\pi \pm \frac{\pi}{4}$ , so

$x = 2n\pi + \frac{5\pi}{12} \quad \text{or} \quad x = 2n\pi - \frac{\pi}{12}$

4. Equations with no solution (use ranges)

$\sin x + \cos x = 2$ has no solution, because $\sin x + \cos x \le \sqrt{2}$ . Similarly, $\sin x = \frac{\pi}{2}$ has no solution because $\frac{\pi}{2} > 1$ . Checking ranges first can save a lot of algebra.

Counting solutions in an interval

List the general solution, then substitute $n = 0, \pm 1, \pm 2, \ldots$ and keep the values inside the interval.

Example: How many solutions does $\sin 2x = \frac{1}{2}$ have in $[0, 2\pi]$ ?

Step 1: $2x = n\pi + (-1)^n\frac{\pi}{6}$ , so $x = \frac{n\pi}{2} + (-1)^n\frac{\pi}{12}$ .

Step 2: $n = 0, 1, 2, 3$ give $\frac{\pi}{12}, \frac{5\pi}{12}, \frac{13\pi}{12}, \frac{17\pi}{12}$ . $n = 4$ gives $\frac{25\pi}{12} > 2\pi$ .

Answer: 4 solutions.

A quick check: $\sin 2x$ completes 2 full cycles in $[0, 2\pi]$ , and each cycle crosses the level $\frac{1}{2}$ twice.

Summary table

Equation type	Method
One function	Reduce to a polynomial in that function; reject out-of-range roots
Sums of sines/cosines	Sum-to-product, then set each factor to zero
a cos x + b sin x = c	Divide by $\sqrt{a^2 + b^2}$ ; solution only if $\lvert c\rvert \le \sqrt{a^2 + b^2}$
Squared forms	$\sin^2 x = \sin^2\alpha \Rightarrow x = n\pi \pm \alpha$
Counting roots	Substitute integer values of $n$ and keep those in range

Exam tips

Always state " $n \in \mathbb{Z}$ " with a general solution.
Never divide by $\sin x$ or $\cos x$ ; factor instead, or you may lose solutions.
Squaring both sides can create extra roots; check them in the original equation.
For counting questions, a sketch of the graph often gives the answer faster than listing.

See the NIMCET Trigonometry chapter for related articles.

Common mistakes

Writing $x = 2n\pi \pm \alpha$ for $\sin x = \sin\alpha$ . That form belongs to cosine.
Keeping impossible roots such as $\sin x = 2$ .
Cancelling a common factor like $\cos x$ and losing the solutions $x = (2n + 1)\frac{\pi}{2}$ .
Forgetting that $\tan x = \tan\alpha$ excludes points where $\tan x$ is undefined.

Frequently Asked Questions

What is the general solution of sin x = sin α?

x = nπ + (−1)ⁿα, where n is any integer.

What is the general solution of cos x = cos α?

x = 2nπ ± α, where n is any integer.

What is the general solution of tan x = tan α?

x = nπ + α, where n is any integer.

When does a cos x + b sin x = c have a solution?

Only when |c| ≤ √(a² + b²). Otherwise the equation has no real solution.

What is the difference between principal and general solutions?

Principal solutions are the solutions in [0, 2π). The general solution is a formula, using an integer n, that gives every solution.