Question

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

Answer 1

Hint: Here we can turn the above-given equation into the quadratic equation in $\sin x$ and then we can find the general value of $\sin x$ 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 $x$
So here we are given the equation as:
$\cos 2x + \sin x = 0$$ - - - (1)$
We know that $\cos 2x = {\cos ^2}x - {\sin ^2}x$$ - - - - - (2)$
Also we know that
$
  {\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 ^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 - 2{\sin ^2}x + \sin x = 0 \\
   - 2{\sin ^2}x + \sin x + 1 = 0 \\
 $
Now we get the quadratic equation in $\sin x$ as:
$2{\sin ^2}x - \sin x - 1 = 0$
Now we can write in above equation that $\left( { - \sin x} \right) = \left( { - 2\sin x + \sin x} \right)$
We will get:
$2{\sin ^2}x - 2\sin x + \sin x - 1 = 0$
Simplifying it further we will get:
$2\sin x(\sin x - 1) + (\sin x - 1) = 0$
$\left( {2\sin x + 1} \right)\left( {\sin x - 1} \right) = 0$
So we can say either $\sin x = - \dfrac{1}{2}{\text{ or }}\sin x = 1$
Now we can say that two value of the trigonometric function are $ - \dfrac{1}{2},1$
So value of $x$ will be ${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right){\text{ or }}{\sin ^{ - 1}}\left( 1 \right)$
So this is just the general value of the $x$ we have obtained but there can be many value of $x$ as $\sin x = - \dfrac{1}{2}{\text{ or }}\sin x = 1$ at many values of $x$
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 $x$ in the particular interval. Then we can find the particular values of $x$ but here we need to give the answer as the general value which is ${\sin ^{ - 1}}\left( {\dfrac{1}{2}} \right){\text{ or }}{\sin ^{ - 1}}\left( 1 \right)$.