If $4 \sin ^{-1} x+6 \cos ^{-1} x=3 \pi$, where $-1 \leq x \leq 1$, then $x=$

If $4 \sin ^{-1} x+6 \cos ^{-1} x=3 \pi$, where $-1 \leq x \leq 1$, then $x=$
  1. $\frac{1}{2}$
  2. $\frac{1}{\sqrt{2}}$
  3. $\frac{-1}{2}$
  4. 0

Solution

$\begin{aligned} & 4 \sin ^{-1} x+6 \cos ^{-1} x=3 \pi \\ & \therefore 4\left(\sin ^{-1} x+\cos ^{-1} x\right)+2 \cos ^{-1} x=3 \pi \\ & \therefore 4\left(\frac{\pi}{2}\right)+2 \cos ^{-1} x=3 \pi \quad \Rightarrow 2 \cos ^{-1} x=\pi \\ & \therefore \cos ^{-1} x=\frac{\pi}{2} \Rightarrow x=\cos \frac{\pi}{2}=0\end{aligned}$

Asked in: MHT CET 2021 (20 Sep Shift 1)

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