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

Solve the equation:
sin⁡x+cos⁡x=sin⁡2x−1\sin x + \cos x = \sin 2x - 1

Explanation

Solution

We need the value of x in such a way that both the sides are equated. If the above example is considered, we need to solve the equation using the standard trigonometric formulae, which are as follows:
cos⁡2x+sin⁡2x=1{\cos ^2}x + {\sin ^2}x = 1
2×sin⁡x×cos⁡x=sin⁡2x2 \times \sin x \times \cos x =\sin 2x
Also, we must know the value of sin and cos at particular angles. In this sum we will come across,
sin⁡0=0. Therefore, sin⁡−10=0\sin 0 = 0.\,Therefore,\,{\sin ^{ - 1}}0 = 0
sin⁡π2=1. Therefore, sin⁡−11=π2\sin \dfrac{\pi }{2} = 1.\,Therefore,\,{\sin ^{ - 1}}1 = \dfrac{\pi }{2}
sin⁡π=0. Therefore, sin⁡−10=π\sin \pi = 0.\,Therefore,\,{\sin ^{ - 1}}0 = \pi

Complete step by step answer:
Equation given in the question:
sin⁡x+cos⁡x=sin⁡2x−1\sin x + \cos x = \sin 2x - 1
To find: x
We will start solving the sum by multiplying -1 on both sides. We get,
cos⁡x−sin⁡x=1−sin⁡2x\cos x - \sin x = 1 - \sin 2x
Squaring on both sides, we get
(cos⁡x−sin⁡x)2=(1−sin⁡2x)2{(\cos x - \sin x)^2} = {(1 - \sin 2x)^2}
cos⁡2x+sin⁡2x−2×sin⁡x×cos⁡x=12+sin⁡22x−2×sin⁡2x{\cos ^2}x + {\sin ^2}x - 2 \times \sin x \times \cos x = {1^2} + {\sin ^2}2x - 2 \times \sin 2x
Let us solve the left hand side First, we get,
LHS: cos⁡2x+sin⁡2x−2×sin⁡x×cos⁡x{\cos ^2}x + {\sin ^2}x - 2 \times \sin x \times \cos x
We know that, cos⁡2x+sin⁡2x=1{\cos ^2}x + {\sin ^2}x = 1 --1
And 2×sin⁡x×cos⁡x=sin⁡2x2 \times \sin x \times \cos x = \sin 2x ----2
After replacing equation 1 and 2 in the LHS, we get
cos⁡2x+sin⁡2x−2×sin⁡x×cos⁡x=1−sin⁡2x{\cos ^2}x + {\sin ^2}x - 2 \times \sin x \times \cos x = 1 - \sin 2x
Now, the original equation becomes,
1−sin⁡2x=1+sin⁡22x−2sin⁡2x1 - \sin 2x = 1 + {\sin ^2}2x - 2\sin 2x
Taking all the terms to one side and equating them to zero, we get,
1−sin⁡2x−1−sin⁡22x+2sin⁡2x=01 - \sin 2x - 1 - {\sin ^2}2x + 2\sin 2x = 0
Further solving it and by multiplying with -1, we get, sin⁡22x−sin⁡2x=0{\sin ^2}2x - \sin 2x = 0
sin⁡2x×(sin⁡2x−1)=0\sin 2x \times (\sin 2x - 1) = 0
Therefore, sin⁡2x=0  or (sin⁡2x−1)=0\sin 2x = 0\,\,or\,(\sin 2x - 1) = 0
When,sin⁡2x=0 \sin 2x = 0\,, 2x=0  or 2x=π2x = 0\,\,or\,2x = \pi
I.e. When sin2x=0, we get x= 0 or x=π2x = \,0\,or\,x = \dfrac{\pi }{2} --3
And When,sin⁡2x−1=0 \sin 2x - 1 = 0\,, 2x=π22x = \dfrac{\pi }{2}
I.e. When sin2x=1, we get x=π4x = \dfrac{\pi }{4} --4
Therefore, x=0 or x=π2 or x=π4 x = 0\,or\,x = \dfrac{\pi }{2}\,or\,x = \dfrac{\pi }{4}\, from equations 3 and 4.

Note:
Few things should be kept in mind when we come across such questions. Firstly, addition, subtraction, all these methods are to be performed correctly, most of the mistakes happen because of the wrong sign. Secondly, the concepts of trigonometric equations must be well known. Just to be sure about your answer, you can put the value of x in the original equation and compare. And, also that x can take more than 1 value in such sums. Every value obtained is correct.