Question

Solve the given equation \[2{\sin ^2}\dfrac{x}{3} = 1\] for \[ - \pi \leqslant x \leqslant \pi \].

Answer 1

Hint: First rearrange the given equation in terms of trigonometric function and given variable only. That is trigonometric function along with x on one side and remaining terms on other side Then we will find the value of x from the given range of angles.

Complete step-by-step answer:
Given that,
\[2{\sin ^2}\dfrac{x}{3} = 1\]
Divide both sides by 2.
\[{\sin ^2}\dfrac{x}{3} = \dfrac{1}{2}\]
Taking square root on both sides,
\[\sin \dfrac{x}{3} = \dfrac{1}{{\sqrt 2 }}\]
But we know that
\[\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}\]
Thus,
\[\dfrac{x}{3} = {\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right)\] ic
\[\dfrac{x}{3} = {45^ \circ }\]
But \[\sin {45^ \circ } = \dfrac{\pi }{4}\]
Range provided is \[ - \pi \leqslant x \leqslant \pi \].
Thus,
\[\sin (\pi + \theta ) = - \sin \theta \]
Here, \[\theta = \dfrac{\pi }{4}\]
\[ \Rightarrow \sin (\pi + \dfrac{\pi }{4}) = - \sin \dfrac{\pi }{4}\]
\[ \Rightarrow - \dfrac{1}{{\sqrt 2 }}\]
Thus,
\[
  \dfrac{x}{3} = \dfrac{{ - 1}}{{\sqrt 2 }} \\
   \Rightarrow x = \dfrac{{ - 3}}{{\sqrt 2 }} \\
\]

Note: Here trigonometric function with range is given but a student should know trigonometric ratios of additional angles. Those include \[ + \theta \] or \[ - \theta \] related to all trigonometric functions. Here remember the identity \[\sin (\pi + \theta ) = - \sin \theta \].
Don’t get confused between $sin2\theta $ and 2$sin\theta$. Both are totally different. In earlier the angle is doubled while in later the function is doubled.