Question

How do you solve $ \sin x=\cos 2x-1 $ ?

Answer 1

Hint: In this problem, we need to solve the given equation i.e., we need to calculate the values of $ x $ which satisfies the given equation. We can observe that the given consists of trigonometric ratios $ \sin $ and $ \cos $ . So, we will convert the given equation into a single trigonometric ratio either $ \sin $ or $ \cos $. We have the trigonometric formula $ \cos 2x=1-2{{\sin }^{2}}x $ . We will use this formula and simplify the given equation. After simplifying the given equation, we will equate it to zero and calculate the solution of the given equation.

Complete step by step answer:
Given that, $ \sin x=\cos 2x-1 $ .
We have the trigonometric formula $ \cos 2x=1-2{{\sin }^{2}}x $ . Substituting this value in the above equation, then we will have
 $ \Rightarrow \sin x=1-2{{\sin }^{2}}x-1 $
We know that $ +a-a=0 $ . Now the above equation is modified as
 $ \Rightarrow \sin x=-2{{\sin }^{2}}x $
Adding the term $ 2{{\sin }^{2}}x $ on both sides of the above equation, then we will get
 $ \Rightarrow 2{{\sin }^{2}}x+\sin x=2{{\sin }^{2}}x-2{{\sin }^{2}}x $
Again, applying the formula $ +a-a=0 $ in the above equation, then we will get
 $ \Rightarrow 2{{\sin }^{2}}x+\sin x=0 $
Now we have converted the given equation in terms of $ \sin x $ . To solve the above equation, we will simplify the above equation.
Taking the term $ \sin x $ as common in the above equation, then we will get
 $ \Rightarrow \sin x\left( 2\sin x+1 \right)=0 $
Equating each term in the above equation to zero, then we will get
 $ \sin x=0 $ or $ 2\sin x+1=0\Rightarrow \sin x=-\dfrac{1}{2} $
Now the solutions for the above equations are
 $ x=n\pi $ , $ x=-\dfrac{\pi }{6}+2n\pi $ and $ x=\pi +\dfrac{\pi }{6}+2n\pi $ where $ n $ is any integer.

Note:
 In the problem, they have not mentioned the range in which we need to calculate the solution. So, we have taken the general solution. Sometimes they may ask to calculate the solution in $ 0 $ to $ 2\pi $ then we need to calculate all the values of $ x $ where the given solution is satisfied.