Question

Find all the values of $x$ , if the trigonometric equation $\sin \left( {2x} \right) = 4\cos \left( x \right)$ holds.

Answer 1

Hint : First of all let’s look at the given trigonometric equation. And then we use some suitable formulae we know and solve for the value of $x$ . We finally arrive at a simple trigonometric equation. From that equation, we can easily determine the value of $x$ .

Complete step-by-step answer :
First of all, let’s look at the given trigonometric equation.
$\sin \left( {2x} \right) = 4\cos \left( x \right)$ ,
Let’s use the suitable formulae, the suitable formula is
$\sin \left( {2x} \right) = 2\sin x\cos x$
Let’s apply the above formula in the given trigonometric equation, we get
$\sin \left( {2x} \right) = 2\sin x\cos x = 4\cos x$
$\Rightarrow 2\sin x\cos x = 4\cos x$ ,
Let’s divide the whole equation by 2, we get
$\Rightarrow \sin x\cos x = 2\cos x$ ,
Let’s take all the expressions into one side of the equation.
$\Rightarrow \sin x\cos x - 2\cos x = 0$ ,
Let's use the $\cos x$ term in both expressions.
$\Rightarrow \cos x\left( {\sin x - 2} \right) = 0$ ,
We know that, if $ab = 0 \Rightarrow a = 0$ or $b = 0$ , so by applying this algorithm to the above equation, we get
$\Rightarrow \cos x = 0$ or $\sin x - 2 = 0$ ,
On further simplifications, we get
$\Rightarrow \cos x = 0$ or $\sin x = 2$
But we know that the value of $\sin x$ lies between $- 1$ and $1$ .
Therefore there is no $x$ such that $\sin x = 2$
Now we left only with the equation, that is
$\cos x = 0$ ,
We have the values of x when $\cos x = 0$ is $x = 2n\pi \pm \dfrac{\pi }{2}$ .
So, The values of $x$ that satisfy the trigonometric equation $\sin \left( {2x} \right) = 4\cos \left( x \right)$ are the same as the equation $\cos x = 0$ , that is $x = 2n\pi \pm \dfrac{\pi }{2}$ .

Note : Observe the given trigonometric equation $\sin \left( {2x} \right) = 4\cos \left( x \right)$ everyone tries to cancel out the term $\cos x$ after some simplification. But we should not cancel them when it is equal to $0$ . Since we cancel the $\cos x$ term we will leave out only with the term $\sin x = 2$ which has no solutions. So, we must be careful while canceling the terms not only in the trigonometry, we should also check the term whether it can be equal to $0$ or not while we are going to cancel it.