Question

How do you solve $4\sin x\cos x=1$?

Answer 1

This type of question is based on the concept of trigonometry. Here, simplify the given function using the trigonometric identity $2\sin x\cos x=\sin 2x$. And divide the whole equation by 2 and cancel out the common terms. Then, use trigonometric inverse identity, that is, ${{\sin }^{-1}}\left( \sin \theta \right)=\theta$ and simplify the given function. We know that, $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$. Using this, we can find the value of x.

Complete answer:
According to the question, we are asked to find the value of $4\sin x\cos x=1$.
We have been given the function is $4\sin x\cos x=1$. --------(1)
Consider the R.H.S first.
$\Rightarrow 4\sin x\cos x=2\times 2\sin x\cos x$
We know that, $2\sin x\cos x=\sin 2x$.
Using the trigonometric identity $2\sin x\cos x=\sin 2x$, we get,
$4\sin x\cos x=2\sin 2x$
Therefore,
$2\sin 2x=1$
On further simplification, we get,
$\sin 2x=\dfrac{1}{2}$ -----------(2)
Now, take ${{\sin }^{-1}}$ on both the sides of the equation (2).
${{\sin }^{-1}}\left( \sin 2x \right)={{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ -------(3)
We know that, ${{\sin }^{-1}}\left( \sin \theta \right)=\theta$.
Using this identity in equation (3), we get,
$2x={{\sin }^{-1}}\left( \dfrac{1}{2} \right)$
$\therefore x=\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ -------(4)
We know that $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$.
$\Rightarrow {{\sin }^{-1}}\left[ \sin \left( \dfrac{\pi }{6} \right) \right]={{\sin }^{-1}}\left( \dfrac{1}{2} \right)$
$\Rightarrow {{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{6}$
But the same value repeats after an interval $2\pi$ in a trigonometric cycle.
Therefore,
${{\sin }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{6}+2n\pi$ , where n is natural numbers.
We get the value of ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ on putting the value of n as 0,1,2 and so on.
Substitute the obtained value of ${{\sin }^{-1}}\left( \dfrac{1}{2} \right)$ in equation (4).
$x=\dfrac{1}{2}\left( \dfrac{\pi }{6}+2n\pi \right)$
On further simplifications, we get,
$\Rightarrow x=\dfrac{\pi }{6\times 2}+\dfrac{2n\pi }{2}$
$\therefore x=\dfrac{\pi }{12}+n\pi$
Hence, the value of x in the function $4\sin x\cos x=1$ is $\dfrac{\pi }{12}+n\pi$ .

Note: We must be thorough with the trigonometric identities. We should avoid calculation mistakes based on sign conventions. It is advisable to first simplify the given equation and then solve to avoid confusion. Trigonometric identities for inverse should also be used to find the value of x. We can also simplify the given equation by cross- multiplying the given function.