Question

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

Answer 1

Hint: 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:
${\cos ^2}x + {\sin ^2}x = 1$
\[2 \times \sin x \times \cos x =\sin 2x\]
Also, we must know the value of sin and cos at particular angles. In this sum we will come across,
$\sin 0 = 0.\,Therefore,\,{\sin ^{ - 1}}0 = 0$
$\sin \dfrac{\pi }{2} = 1.\,Therefore,\,{\sin ^{ - 1}}1 = \dfrac{\pi }{2}$
$\sin \pi = 0.\,Therefore,\,{\sin ^{ - 1}}0 = \pi $

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