Solving trigonometric equations

One equation, many solutions

In Class 10 an equation such as $\sin x = \dfrac{1}{2}$ had one answer, $30^\circ$. Now $x$ can be any real number. The sine wave rises and falls for ever, so a level line at height $\dfrac{1}{2}$ crosses it again and again.

• A principal solution is a solution with $0 \le x < 2\pi$.
• The general solution is one formula, with an integer $n$, that gives every solution.

For $\sin x = \dfrac{1}{2}$ the principal solutions are $\dfrac{\pi}{6}$ and $\dfrac{5\pi}{6}$. Adding any number of whole turns gives the rest.

The three general solutions

Let $\alpha$ be one solution, usually the principal value from a calculator. Then, for any integer $n$:

• $\sin x = \sin\alpha$ gives $x = n\pi + (-1)^n\alpha$;
• $\cos x = \cos\alpha$ gives $x = 2n\pi \pm \alpha$;
• $\tan x = \tan\alpha$ gives $x = n\pi + \alpha$.

Why sine has this form. Two points of the unit circle have the same height: the one at $\alpha$ and its mirror image at $\pi - \alpha$. With whole turns added, the solutions are $2m\pi + \alpha$ and $(2m + 1)\pi - \alpha$. An even $n$ gives the first family and an odd $n$ the second, so both fit $n\pi + (-1)^n\alpha$.

Why cosine has this form. The points at $\alpha$ and $-\alpha$ have the same x-coordinate. Whole turns give $2n\pi \pm \alpha$.

Why tangent has this form. Opposite points, at $\alpha$ and $\alpha + \pi$, have the same ratio $\dfrac{y}{x}$. So tangent repeats every $\pi$.

Three special cases come up again and again.

• $\sin x = 0$ gives $x = n\pi$.
• $\cos x = 0$ gives $x = (2n + 1)\dfrac{\pi}{2}$.
• $\sin^2 x = \sin^2\alpha$, $\cos^2 x = \cos^2\alpha$ and $\tan^2 x = \tan^2\alpha$ all give $x = n\pi \pm \alpha$.

Example 1: one value of sine

Solve $2\sin x + \sqrt{2} = 0$.

1. Make the function alone. $\sin x = -\dfrac{1}{\sqrt{2}}$.
2. Find one angle. $\sin\left(-\dfrac{\pi}{4}\right) = -\dfrac{1}{\sqrt{2}}$, so $\alpha = -\dfrac{\pi}{4}$.
3. The general solution. $x = n\pi + (-1)^n\left(-\dfrac{\pi}{4}\right)$.
4. The principal solutions. $n = 1$ gives $\pi + \dfrac{\pi}{4} = \dfrac{5\pi}{4}$, and $n = 2$ gives $2\pi - \dfrac{\pi}{4} = \dfrac{7\pi}{4}$.

Both lie in quadrants III and IV, where sine is negative. That is a quick check.

Set $k = 0.5$, then $k = -0.7$. The dots move, but they always come in the pattern $\alpha$, $\pi - \alpha$, $2\pi + \alpha$, and so on.

Equations that factorise

Many equations hold more than one function. Use an identity to bring them to one function, or factorise.

Example 2: a quadratic in $\cos x$. Solve $2\sin^2 x + 3\cos x = 0$.

1. One function. $\sin^2 x = 1 - \cos^2 x$, so $2 - 2\cos^2 x + 3\cos x = 0$, that is $2\cos^2 x - 3\cos x - 2 = 0$.
2. Factorise. $(2\cos x + 1)(\cos x - 2) = 0$.
3. Reject the impossible root. $\cos x = 2$ has no solution, since $|\cos x| \le 1$.
4. Solve. $\cos x = -\dfrac{1}{2} = \cos\dfrac{2\pi}{3}$, so $x = 2n\pi \pm \dfrac{2\pi}{3}$.

Example 3: a sum turned into a product. Solve $\cos x + \cos 3x = 0$.

1. Sum to product. $\cos 3x + \cos x = 2\cos 2x\cos x$.
2. Two factors. $\cos 2x = 0$ gives $2x = (2n + 1)\dfrac{\pi}{2}$, so $x = (2n + 1)\dfrac{\pi}{4}$. And $\cos x = 0$ gives $x = (2n + 1)\dfrac{\pi}{2}$.
3. In one turn. $\dfrac{\pi}{4}$, $\dfrac{3\pi}{4}$, $\dfrac{5\pi}{4}$ and $\dfrac{7\pi}{4}$ from the first factor, and $\dfrac{\pi}{2}$ and $\dfrac{3\pi}{2}$ from the second: six solutions.

Check every root

• Dividing loses roots. From $\sin^2 x = \sin x$, dividing by $\sin x$ leaves only $\sin x = 1$. The roots with $\sin x = 0$ are lost. Write $\sin x(\sin x - 1) = 0$ instead.
• Squaring adds roots. Squaring $\sin x + \cos x = 1$ gives $1 + \sin 2x = 1$, so $x = \dfrac{n\pi}{2}$. In one turn that gives $0$, $\dfrac{\pi}{2}$, $\pi$ and $\dfrac{3\pi}{2}$. Only $0$ and $\dfrac{\pi}{2}$ satisfy the first equation.
• Undefined values. If the equation holds $\tan x$ or $\sec x$, throw away any root where $\cos x = 0$.

Practise with new numbers

Try it in Python

This program lists the solutions in one turn of $\sin x = v$, $\cos x = v$ and $\tan x = v$, from the general solutions. Then it puts each one back into its equation.

Common mistakes

• Sine with $2n\pi + \alpha$. That family misses the mirror solutions $\pi - \alpha$. Use $n\pi + (-1)^n\alpha$.
• Cosine with $n\pi \pm \alpha$. The point at $\pi - \alpha$ has the opposite x-coordinate, so it does not solve $\cos x = \cos\alpha$.
• Keeping $\cos x = 2$. A value of sine or cosine outside $[-1, 1]$ gives no solution.
• Dividing by a factor. It loses every root where that factor is zero.
• Mixing units. Never write $n\pi + 30^\circ$. Keep the whole answer in radians.

Remember these

• $\sin x = \sin\alpha$: $x = n\pi + (-1)^n\alpha$.
• $\cos x = \cos\alpha$: $x = 2n\pi \pm \alpha$.
• $\tan x = \tan\alpha$: $x = n\pi + \alpha$.
• Bring the equation to one function, factorise, reject impossible values, and check the roots after squaring.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the principal solutions of $\tan x = -\sqrt{3}$.

2. Write the general solution of $2\cos x + \sqrt{3} = 0$.

3. How many solutions does $\sin x = 0.9$ have with $0 \le x \le 4\pi$?

4. Raj solves $2\cos^2 x = \cos x$ by dividing both sides by $\cos x$. Which solutions does he lose?

5. Solve $\tan 2x = 1$, and give the solutions with $0 \le x < \pi$.