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Question: How do you solve \(2{{\sin }^{2}}x=\cos x+1\) in the interval \(0\le x\le 2\pi \) ?...

How do you solve 2sin⁡2x=cos⁡x+12{{\sin }^{2}}x=\cos x+1 in the interval 0≤x≤2π0\le x\le 2\pi ?

Explanation

Solution

We have been given an inequation consisting of two trigonometric functions, sine of x and cosine of x. Since the mentioned interval is 0≤x≤2π0\le x\le 2\pi , therefore, we shall only find the principal values only for which this equation holds true. Firstly, we shall convert the entire equation in terms of the cosine function using the trigonometric identities. Then we shall solve it further like a quadratic function which would be in terms of cos x.

Complete step by step solution:
Given that 2sin⁡2x=cos⁡x+12{{\sin }^{2}}x=\cos x+1.
We shall use the trigonometric identity sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1 to substitute the square of sine of x with 1 minus the square of cosine of x, that is, sin⁡2x=1−cos⁡2x{{\sin }^{2}}x=1-{{\cos }^{2}}x.
⇒2(1−cos⁡2x)=cos⁡x+1\Rightarrow 2\left( 1-{{\cos }^{2}}x \right)=\cos x+1
⇒2−2cos⁡2x=cos⁡x+1\Rightarrow 2-2{{\cos }^{2}}x=\cos x+1
Transposing the constant term 1 and cosx to the left hand side of equation, we get
⇒2−2cos⁡2x−cos⁡x−1=0\Rightarrow 2-2{{\cos }^{2}}x-\cos x-1=0
⇒1−2cos⁡2x−cos⁡x=0\Rightarrow 1-2{{\cos }^{2}}x-\cos x=0
We will multiply the entire equation with a negative sign.
⇒2cos⁡2x+cos⁡x−1=0\Rightarrow 2{{\cos }^{2}}x+\cos x-1=0
Now, we shall factor the quadratic equation formed above and further group them into two groups.
⇒2cos⁡2x+2cos⁡x−cos⁡x−1=0\Rightarrow 2{{\cos }^{2}}x+2\cos x-\cos x-1=0
⇒2cos⁡x(cos⁡x+1)−1(cos⁡x+1)=0\Rightarrow 2\cos x\left( \cos x+1 \right)-1\left( \cos x+1 \right)=0
⇒(cos⁡x+1)(2cos⁡x−1)=0\Rightarrow \left( \cos x+1 \right)\left( 2\cos x-1 \right)=0
Thus, cos⁡x+1=0\cos x+1=0 and 2cos⁡x−1=02\cos x-1=0.
For cos⁡x+1=0\cos x+1=0,
⇒cos⁡x=−1\Rightarrow \cos x=-1
This holds true for x=πx=\pi in the interval 0≤x≤2π0\le x\le 2\pi .
For 2cos⁡x−1=02\cos x-1=0,
⇒cos⁡x=−12\Rightarrow \cos x=-\dfrac{1}{2}
This holds true for x=2π3,4π3x=\dfrac{2\pi }{3},\dfrac{4\pi }{3} in the interval 0≤x≤2π0\le x\le 2\pi .

Therefore, the solution of 2sin⁡2x=cos⁡x+12{{\sin }^{2}}x=\cos x+1 in the interval 0≤x≤2π0\le x\le 2\pi is x=2π3,π,4π3x=\dfrac{2\pi }{3},\pi ,\dfrac{4\pi }{3}.

Note: In order to find the solution of various trigonometric equations, we must have prior knowledge of the main trigonometric identities. Also, we could have substituted the quadratic equation formed in terms of cosine of x function with any variable-t to proceed with solving the quadratic equation. Later we could equate the calculated values of t-variable equal to cos x to find our final solution.