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Question: Solve: \(1 - \sin 2x = \cos x - \sin x\)...

Solve: 1−sin⁡2x=cos⁡x−sin⁡x1 - \sin 2x = \cos x - \sin x

Explanation

Solution

To solve 1−sin⁡2x=cos⁡x−sin⁡x1 - \sin 2x = \cos x - \sin x, first of all square the given equation on both sides. Then simplify and use the formula 2sin⁡xcos⁡x=sin⁡2x2\sin x\cos x = \sin 2x and then gather the like terms on one side and then you will get the answer.

Complete step-by-step answer:
In this question, we are given a trigonometric equation and we have to find its solution.
The equation is: 1−sin⁡2x=cos⁡x−sin⁡x1 - \sin 2x = \cos x - \sin x- - - - - - - - - (1)
Now, to solve this equation we have to use relations, formulas or even mathematical operations.
First of all, we are going to square both the sides of equation (1). So, on squaring equation (1), we get
→(1−sin⁡2x)2=(cos⁡x−sin⁡x)2\to {\left( {1 - \sin 2x} \right)^2} = {\left( {\cos x - \sin x} \right)^2}- - - - - - - - - - - (2)
Now, we know the formula of (a−b)2{\left( {a - b} \right)^2}.
→(a−b)2=a2−2ab+b2\to {\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}
Here, on the LHS a=1a = 1 and b=sin⁡2xb = \sin 2x and on the RHS, a=cos⁡xa = \cos x and b=sin⁡xb = \sin x.
Therefore, equation (2) will become
→1−2sin⁡2x+sin⁡22x=cos⁡2x−2sin⁡xcos⁡x+sin⁡2x\to 1 - 2\sin 2x + {\sin ^2}2x = {\cos ^2}x - 2\sin x\cos x + {\sin ^2}x
→1−2sin⁡2x+sin⁡22x=sin⁡2x+cos⁡2x−2sin⁡xcos⁡x\to 1 - 2\sin 2x + {\sin ^2}2x = {\sin ^2}x + {\cos ^2}x - 2\sin x\cos x- - - - - - - - - - (3)
Now, we know the sin⁡2x+cos⁡2x=1{\sin ^2}x + {\cos ^2}x = 1 and 2sin⁡xcos⁡x=sin⁡2x2\sin x\cos x = \sin 2x. So, substituting these values in equation (3), we get
→1−2sin⁡2x+sin⁡22x=1−sin⁡2x\to 1 - 2\sin 2x + {\sin ^2}2x = 1 - \sin 2x- - - - - - - - (4)
Now, gather sin2x terms on one side of equation and constants on the other side. Therefore, equation (4) becomes
→sin⁡22x−2sin⁡2x+sin⁡2x=1−1\to {\sin ^2}2x - 2\sin 2x + \sin 2x = 1 - 1
→sin⁡22x−sin⁡2x=0\to {\sin ^2}2x - \sin 2x = 0- - - - - - (5)
Now, we can take sin2x common. Therefore, equation (5) becomes
→sin⁡2x(sin⁡2x−1)=0\to \sin 2x\left( {\sin 2x - 1} \right) = 0

Therefore,
→sin⁡2x=0 →x=0  \to \sin 2x = 0 \\\ \to x = 0 \\\
OR
→sin⁡2x−1=0 →sin⁡2x=1 →2x=π2 →x=π4  \to \sin 2x - 1 = 0 \\\ \to \sin 2x = 1 \\\ \to 2x = \dfrac{\pi }{2} \\\ \to x = \dfrac{\pi }{4} \\\

Note: Some of the important trigonometric relations one should always keep in mind are:
sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x
cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = {\cos ^2}x - {\sin ^2}x
cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2{\sin ^2}x
cos⁡2x=2cos⁡2x−1\cos 2x = 2{\cos ^2}x - 1
cos⁡2x=1−tan⁡2x1+tan⁡2x\cos 2x = \dfrac{{1 - {{\tan }^2}x}}{{1 + {{\tan }^2}x}}