Question

How do you solve $\cos 2x+2\sin x=0$ using the double angle identity?

Answer 1

Hint: In this question, we have to find the value of x. It is given that the equation consists of the trigonometric functions, so we solve this problem using the double angle identity. We start solving this problem by applying the double angle identity $\cos 2a=1-2{{\sin }^{2}}a$ in the equation, and get the quadratic equation. Then, we will use the discriminant formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ in the equation and make the necessary calculations. After that, we get two new equations and solve them separately to get the value of x, which is our required answer.

Complete answer:
According to the question, we have to find the value of x of the trigonometric equations
So, we solve this problem using the double angle identity.
Equation: $\cos 2x+2\sin x=0$ -------------- (1)
First, we will apply the trigonometric double-angle identity $\cos 2a=1-2{{\sin }^{2}}a$ in equation (1), we get
$\Rightarrow 1-2{{\sin }^{2}}x+2\sin x=0$
$\Rightarrow -2{{\sin }^{2}}x+2\sin x+1=0$ ----------- (2)
Now, we see equation (2) is in a form of a quadratic equation $a{{x}^{2}}+x+c=0$ , therefore we will find the value of sinx, using discriminant formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ --------- (3)
Now, we will find the value of a, b, and c by comparing the general form of equation and equation (2), we get
$a=-2\text{, }b=2\text{, and }c=1$ , therefore we will now put these value in the equation (3), we get
$\Rightarrow \sin x=\dfrac{-2\pm \sqrt{{{2}^{2}}-4.(-2).(1)}}{2.(-2)}$
On further the above equation, we get
$\begin{align}
  & \Rightarrow \sin x=\dfrac{-2\pm \sqrt{4+8}}{-4} \\
 & \Rightarrow \sin x=\dfrac{-2\pm \sqrt{12}}{-4} \\
\end{align}$
Now, we know that $\sqrt{12}=2\sqrt{3}$ , therefore we get
$\Rightarrow \sin x=\dfrac{-2\pm 2\sqrt{3}}{-4}$
Now, we will take 2 common in the numerator, we get
$\Rightarrow \sin x=\dfrac{2(-1\pm \sqrt{3})}{-4}$
On further solving, we get
$\Rightarrow \sin x=\dfrac{(-1\pm \sqrt{3})}{-2}$
So, now we will split $\pm $ sign in the above equation, we get
$\Rightarrow \sin x=\dfrac{(-1+\sqrt{3})}{-2}$ --------- (4) or
$\Rightarrow \sin x=\dfrac{(-1-\sqrt{3})}{-2}$ ----------- (5)
Now, first, we will solve equation (4), we get
$\begin{align}
  & \Rightarrow \sin x=\dfrac{-1}{2}-\dfrac{\sqrt{3}}{2} \\
 & \Rightarrow x={{\sin }^{-1}}\left( \dfrac{-1}{2}-\dfrac{\sqrt{3}}{2} \right) \\
\end{align}$
Now, we will solve equation (5), we get
$\begin{align}
  & \Rightarrow \sin x=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2} \\
 & \Rightarrow x={{\sin }^{-1}}\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2} \right) \\
\end{align}$
Therefore, for the equation $\cos 2x+2\sin x=0$ using the double angle identity, we get two values of x, that is $x={{\sin }^{-1}}\left( \dfrac{-1}{2}-\dfrac{\sqrt{3}}{2} \right)$ and $\Rightarrow x={{\sin }^{-1}}\left( \dfrac{1}{2}+\dfrac{\sqrt{3}}{2} \right)$ , which is our required answer to the problem.

Note: While solving this problem, always mention the identity you are using in each step. Since in this question you have to solve this problem by using double-angle identity only, so do not use any other identity to solve this problem. Also, when you get two separate equations while solving this problem, solve both the equations separately, so not you will not get confused and get the correct answer.