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

Solve the given equation 2sin⁡2x3=12{\sin ^2}\dfrac{x}{3} = 1 for −π⩽x⩽π- \pi \leqslant x \leqslant \pi.

Explanation

Solution

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,
2sin⁡2x3=12{\sin ^2}\dfrac{x}{3} = 1
Divide both sides by 2.
sin⁡2x3=12{\sin ^2}\dfrac{x}{3} = \dfrac{1}{2}
Taking square root on both sides,
sin⁡x3=12\sin \dfrac{x}{3} = \dfrac{1}{{\sqrt 2 }}
But we know that
sin⁡45∘=12\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}
Thus,
x3=sin⁡−1(12)\dfrac{x}{3} = {\sin ^{ - 1}}\left( {\dfrac{1}{{\sqrt 2 }}} \right) ic
x3=45∘\dfrac{x}{3} = {45^ \circ }
But sin⁡45∘=π4\sin {45^ \circ } = \dfrac{\pi }{4}
Range provided is −π⩽x⩽π- \pi \leqslant x \leqslant \pi.
Thus,
sin⁡(π+θ)=−sin⁡θ\sin (\pi + \theta ) = - \sin \theta
Here, θ=π4\theta = \dfrac{\pi }{4}
⇒sin⁡(π+π4)=−sin⁡π4\Rightarrow \sin (\pi + \dfrac{\pi }{4}) = - \sin \dfrac{\pi }{4}
⇒−12\Rightarrow - \dfrac{1}{{\sqrt 2 }}
Thus,

x3=−12 ⇒x=−32  \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 +θ+ \theta or −θ- \theta related to all trigonometric functions. Here remember the identity sin⁡(π+θ)=−sin⁡θ\sin (\pi + \theta ) = - \sin \theta .
Don’t get confused between sin2θsin2\theta and 2sinθsin\theta. Both are totally different. In earlier the angle is doubled while in later the function is doubled.