TS EAMCET · Maths · Inverse Trigonometric Functions
If \(\cos ^{-1} 2 x+\cos ^{-1} 3 x=\frac{\pi}{3}\), then \(x=\)
- A \(\frac{\sqrt{3}}{2 \sqrt{7}}\)
- B \(\frac{\sqrt{3}}{\sqrt{7}}\)
- C \(\frac{\sqrt{2}}{\sqrt{5}}\)
- D \(\frac{\sqrt{3}}{2 \sqrt{5}}\)
Answer & Solution
Correct Answer
(A) \(\frac{\sqrt{3}}{2 \sqrt{7}}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \cos ^{-1} 2 x+\cos ^{-1} 3 x=\frac{\pi}{3} \\ & \cos ^{-1}\left\{2 x \cdot 3 x-\sqrt{1-4 x^2} \sqrt{1-9 x^2}\right\}=\frac{\pi}{3} \\ & \Rightarrow 6 x^2-\sqrt{1-4 x^2} \sqrt{1-9 x^2}=\frac{1}{2} \\ & \Rightarrow-\sqrt{1-4 x^2} \sqrt{1-9 x^2}=\frac{1}{2}-6 x^2…
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