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Question: How do you solve \(2{{\cos }^{2}}x-2{{\sin }^{2}}x=1\) and find all solutions in the interval \(0\le...

How do you solve 2cos⁡2x−2sin⁡2x=12{{\cos }^{2}}x-2{{\sin }^{2}}x=1 and find all solutions in the interval 0≤x<3600\le x<360 ?

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<3600\le x<360, therefore, we shall only find the principal values only for which this equation holds true. Firstly, we shall take 2 common terms on the left hand side of the equation and then we shall use trigonometric properties to find the final solution.

Complete step by step solution:
Given that 2cos⁡2x−2sin⁡2x=12{{\cos }^{2}}x-2{{\sin }^{2}}x=1.
We shall take 2 common from the terms in the left hand side of the equation.
⇒2(cos⁡2x−sin⁡2x)=1\Rightarrow 2\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)=1
Dividing both sides by 2, we get
⇒cos⁡2x−sin⁡2x=12\Rightarrow {{\cos }^{2}}x-{{\sin }^{2}}x=\dfrac{1}{2}
Now, we shall use a trigonometric property where the square of cosine of x minus the square of sine of x equal to cosine of twice of x, that is, cos⁡2x−sin⁡2x=cos⁡2x{{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x. Substituting the value of cos⁡2x−sin⁡2x{{\cos }^{2}}x-{{\sin }^{2}}x from this identity, we get
⇒cos⁡2x=12\Rightarrow \cos 2x=\dfrac{1}{2}
We know that cos⁡π3=12\cos \dfrac{\pi }{3}=\dfrac{1}{2} and cos⁡5π3=12\cos \dfrac{5\pi }{3}=\dfrac{1}{2}
Therefore, 2x=π32x=\dfrac{\pi }{3} and 2x=5π32x=\dfrac{5\pi }{3}
Dividing 2 from both sides of this equation, we get
⇒x=π6\Rightarrow x=\dfrac{\pi }{6} and x=5π6x=\dfrac{5\pi }{6}
Thus, x=π6,5π6x=\dfrac{\pi }{6},\dfrac{5\pi }{6}.
Therefore, the solution of 2cos⁡2x−2sin⁡2x=12{{\cos }^{2}}x-2{{\sin }^{2}}x=1 in the interval 0≤x<3600\le x <360 is x=π6,5π6x=\dfrac{\pi }{6},\dfrac{5\pi }{6}.

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 used another trigonometric property, 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. Using this a quadratic equation would have formed in terms of sine of x which could be solved further to get the final solution.