Question

How do you find the exact value of the equation $\sin 4x=-2\sin 2x$ in the interval $\left[ 0,2\pi \right)$?

Answer 1

In the given question, we have been asked to find the value of the given equation for the given interval i.e. $\left[ 0,2\pi \right)$. In order to solve the question, first we need to simplify the equation by substituting by using the trigonometric identity i.e. $\sin 2a=2\sin a\cos a$. Then solving each factor by considering the interval and you will get the exact values of ‘x’.

Complete step by step solution:
We have given that,
$\sin 4x=-2\sin 2x$
Rewrite the equation as,
$\sin 4x+2\sin 2x=0$
As using the trigonometric identity i.e. $\sin 2a=2\sin a\cos a$
Here, a = 2x then 2a = 4x.
Substituting the value of$\sin 4x=2\sin 2x\cos 2x$ in the given above expression, we get
$2\sin 2x\cos 2x+2\sin 2x=0$
Taking out the common factor from the above expression, we get
$2\sin 2x\left( \cos 2x+1 \right)=0$
Now solving,
$\Rightarrow \sin 2x=0$
On a trigonometric unit circle;
Unit circle of the trigonometric given 3 solutions for 2x i.e.
$\Rightarrow 2x=0\Rightarrow x=0$
$\Rightarrow 2x=\pi \Rightarrow x=\dfrac{\pi }{2}$
$\Rightarrow 2x=2\pi \Rightarrow x=\pi$
Now solving,
$\Rightarrow \cos 2x+1=0$
Rewritten as,
$\Rightarrow \cos 2x=-1$
On a trigonometric unit circle;
Unit circle of the trigonometric given 2 solutions for 2x i.e.
$\Rightarrow 2x=\pi \Rightarrow x=\dfrac{\pi }{2}$
$\Rightarrow 2x=3\pi \Rightarrow x=\dfrac{3\pi }{2}$

Therefore,
The values of $x$ are $0,\dfrac{\pi }{2},\pi ,\dfrac{3\pi }{2}$.

Note: In order to solve these types of questions, you should always need to remember the properties of trigonometric and the trigonometric ratios as well. It will make questions easier to solve. It is preferred that while solving these types of questions we should carefully examine the pattern of the given function and then you would apply the formulas according to the pattern observed. As if you directly apply the formula it will create confusion ahead and we will get the wrong answer.