Solving equations

What you will be able to do

• solve an equation with one inverse function, and see when it has no solution;
• solve an equation with several inverse functions by an identity or by taking a sine, a cosine or a tangent;
• reject a false root by checking the range and the domain.

One inverse function

If the equation is $\sin^{-1}u = a$, then $u = \sin a$, provided that $a$ is in the range of $\sin^{-1}$. If $a$ is outside $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$, there is no solution at all.

• $\sin^{-1}(2x) = \dfrac{\pi}{6}$ gives $2x = \dfrac{1}{2}$ and $x = \dfrac{1}{4}$. Check: $2x = \dfrac{1}{2}$ is in $[-1, 1]$.
• $\sin^{-1}x = 2$ has no solution, because $2$ is larger than $\dfrac{\pi}{2} \approx 1.57$.
• $\cos^{-1}x = -1$ has no solution, because $\cos^{-1}x$ is never negative.

Several inverse functions

There are three ways to remove them.

• Combine with an identity. For example, replace $\cos^{-1}x$ by $\dfrac{\pi}{2} - \sin^{-1}x$. Then there is only one inverse function left.
• Isolate one inverse function and take a sine, cosine or tangent of both sides.
• Use the sum rule for $\tan^{-1}x + \tan^{-1}y$, remembering its condition.

Taking a sine or a tangent loses information, because many angles have the same sine or tangent. So it can create false roots: solutions of the new equation that do not solve the original one.

The steps are always the same.

1. Find the domain of every inverse function in the equation.
2. Take the function of both sides, and solve the algebra.
3. Check every root in the original equation. Throw away any that fail.

A worked example

Solve (a) $\tan^{-1}x + \tan^{-1}2x = \dfrac{\pi}{4}$ and (b) $\cos^{-1}x = 2\sin^{-1}x$.

1. (a) Take the tangent of both sides: $\dfrac{x + 2x}{1 - 2x^2} = 1$. So $3x = 1 - 2x^2$, that is $2x^2 + 3x - 1 = 0$, and $x = \dfrac{-3 \pm \sqrt{17}}{4}$. The two roots are about $0.281$ and $-1.781$. Check. For $x \approx 0.281$ the product $x \cdot 2x \approx 0.16 < 1$, so the sum rule applies and the sum is $\tan^{-1}1 = \dfrac{\pi}{4}$. For $x \approx -1.781$ the product is about $6.3 > 1$ and $x < 0$, so the sum is $-\pi + \dfrac{\pi}{4} = -\dfrac{3\pi}{4}$. That root is false. The only solution is $x = \dfrac{-3 + \sqrt{17}}{4}$.
2. (b) The domain is $[-1, 1]$. Replace $\cos^{-1}x$ by $\dfrac{\pi}{2} - \sin^{-1}x$. Then $\dfrac{\pi}{2} - \sin^{-1}x = 2\sin^{-1}x$, so $\sin^{-1}x = \dfrac{\pi}{6}$, and $x = \dfrac{1}{2}$. Check. $\cos^{-1}\dfrac{1}{2} = \dfrac{\pi}{3}$ and $2\sin^{-1}\dfrac{1}{2} = \dfrac{\pi}{3}$.

Practise with new numbers

Try it in Python

The program solves $\tan^{-1}x + \tan^{-1}2x = \dfrac{\pi}{4}$ by taking the tangent. Then it tests both roots in the original equation.

Common mistakes

• Skipping the check. Taking a tangent or a square can add a root that is false.
• Applying $\sin$ to an impossible value. $\sin^{-1}x = 2$ has no solution, although $\sin 2$ exists.
• Forgetting the domain. A root outside $[-1, 1]$ cannot be used in $\sin^{-1}x$ or $\cos^{-1}x$.
• Using the sum formula without its condition. The product $xy$ decides whether $\pm\pi$ is needed.

Remember these

• $\sin^{-1}u = a$ has a solution only if $a$ is in $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$.
• Replace $\cos^{-1}x$ by $\dfrac{\pi}{2} - \sin^{-1}x$ to reduce the number of inverse functions.
• Taking a sine, cosine or tangent can create false roots. Check every root in the original equation.
• Find the domain first. It can remove a root at once.

Check yourself

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

1. Solve $\tan^{-1}x = \dfrac{\pi}{3}$.

2. Solve $\sin^{-1}x = 2$.

3. Solve $\sin^{-1}x + 3\cos^{-1}x = \dfrac{5\pi}{6}$.

4. Solve $\tan^{-1}x + \tan^{-1}3 = \dfrac{\pi}{2}$.

5. Taking the tangent of $\tan^{-1}x + \tan^{-1}3x = \dfrac{\pi}{4}$ gives $3x^2 + 4x - 1 = 0$, with roots $\dfrac{-2 \pm \sqrt{7}}{3}$. How many of them solve the original equation?

References

• NCERT, Mathematics Textbook for Class XII, Part I, Chapter 2: Inverse Trigonometric Functions.
• OpenStax, Algebra and Trigonometry 2e, Section 8.3: Inverse Trigonometric Functions.
• OpenStax, Calculus Volume 1, Section 1.4: Inverse Functions.