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Question: How do you solve \(\cos 2x + \sin x = 0\)?...

How do you solve cos⁡2x+sin⁡x=0\cos 2x + \sin x = 0?

Explanation

Solution

Here we can turn the above-given equation into the quadratic equation in sin⁡x\sin x and then we can find the general value of sin⁡x\sin x for which it is satisfying and hence we can get the solution of the problem.

Complete step by step solution:
Here we need to find the value of xx
So here we are given the equation as:
cos⁡2x+sin⁡x=0\cos 2x + \sin x = 0 −−−(1) - - - (1)
We know that cos⁡2x=cos⁡2x−sin⁡2x\cos 2x = {\cos ^2}x - {\sin ^2}x −−−−−(2) - - - - - (2)
Also we know that
sin⁡2x+cos⁡2x=1 cos⁡2x=1−sin⁡2x  {\sin ^2}x + {\cos ^2}x = 1 \\\ {\cos ^2}x = 1 - {\sin ^2}x \\\
Now we can substitute this value in the equation (2) and get:
cos⁡2x=1−sin⁡2x−sin⁡2x cos⁡2x=1−2sin⁡2x −−−−−(3)  \cos 2x = 1 - {\sin ^2}x - {\sin ^2}x \\\ \cos 2x = 1 - 2{\sin ^2}x{\text{ }} - - - - - (3) \\\
Now substituting this value we get in equation (3) in the equation (1) we will get:
1−2sin⁡2x+sin⁡x=0 \-2sin⁡2x+sin⁡x+1=0  1 - 2{\sin ^2}x + \sin x = 0 \\\ \- 2{\sin ^2}x + \sin x + 1 = 0 \\\
Now we get the quadratic equation in sin⁡x\sin x as:
2sin⁡2x−sin⁡x−1=02{\sin ^2}x - \sin x - 1 = 0
Now we can write in above equation that (−sin⁡x)=(−2sin⁡x+sin⁡x)\left( { - \sin x} \right) = \left( { - 2\sin x + \sin x} \right)
We will get:
2sin⁡2x−2sin⁡x+sin⁡x−1=02{\sin ^2}x - 2\sin x + \sin x - 1 = 0
Simplifying it further we will get:
2sin⁡x(sin⁡x−1)+(sin⁡x−1)=02\sin x(\sin x - 1) + (\sin x - 1) = 0
(2sin⁡x+1)(sin⁡x−1)=0\left( {2\sin x + 1} \right)\left( {\sin x - 1} \right) = 0
So we can say either sin⁡x=−12 or sin⁡x=1\sin x = - \dfrac{1}{2}{\text{ or }}\sin x = 1
Now we can say that two value of the trigonometric function are −12,1 - \dfrac{1}{2},1
So value of xx will be sin⁡−1(12) or sin⁡−1(1){\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right){\text{ or }}{\sin ^{ - 1}}\left( 1 \right)
So this is just the general value of the xx we have obtained but there can be many value of xx as sin⁡x=−12 or sin⁡x=1\sin x = - \dfrac{1}{2}{\text{ or }}\sin x = 1 at many values of xx
This is also because we are not given any particular interval in which we have to find the solution.

Note:
This problem can also be asked in the manner where we can be asked to find the value of xx in the particular interval. Then we can find the particular values of xx but here we need to give the answer as the general value which is sin⁡−1(12) or sin⁡−1(1){\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right){\text{ or }}{\sin ^{ - 1}}\left( 1 \right).