Question

How do you solve \[\cos x=\sin 2x\sin x\]?

Answer 1

Hint: In this problem, we have to solve and find the value of x from the given trigonometric expression. We should know that to solve these types of problems, we should know some trigonometric formulas, rules, properties to apply and solve. We should also know some sine and cosine degree values to get the exact solution.

Complete step by step answer:
We know that the given expression to be solved for x is,
\[\cos x=\sin 2x\sin x\] ….. (1)
We know that the trigonometric identity,
\[\sin 2x=2\sin x\cos x\]
We can apply the above trigonometric identity in the expression (1), we get
\[\begin{align}
  & \Rightarrow \cos x=2\sin x\cos x\sin x \\
 & \Rightarrow \cos x=2\cos x{{\sin }^{2}}x \\
\end{align}\]
Now we can divide the step by cosine term in order to cancel similar terms, we get
\[\Rightarrow 2{{\sin }^{2}}x=1\]
Now we can divide the number 2 on both the left-hand side and the right-hand side in the above step, we get
\[\Rightarrow {{\sin }^{2}}x=\dfrac{1}{2}\]
We can now take square root 2 on both the left-hand side and the right-hand side in the above step, we get
\[\Rightarrow \sin x=\pm \dfrac{1}{\sqrt{2}}\] .
Here, sin is positive for the first two quadrants and negative for the third and fourth quadrant.
When x is \[\dfrac{\pi }{4}\] and \[\dfrac{3\pi }{4}\], the value of sin x is \[\dfrac{1}{\sqrt{2}}\] and when x is \[\dfrac{5\pi }{4}\] and \[\dfrac{7\pi }{4}\], the value of sin x is \[-\dfrac{1}{\sqrt{2}}\].
Therefore, the value of x = \[\dfrac{\pi }{4},\dfrac{3\pi }{4},\dfrac{5\pi }{4},\dfrac{7\pi }{4}....\]

Note:
Students make mistakes in finding the correct degree value, which should be concentrated. we should know some trigonometric formulas, rules, properties to apply and solve. We should also know some sine and cosine degree values to get the exact solution.