If cos ^{-1} 2x + cos ^{-1} 3x = frac{pi}{3}, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $\cos ^{-1} 2x + \cos ^{-1} 3x = \frac{\pi}{3}$.Using the formula $\cos ^{-1} A + \cos ^{-1} B = \cos ^{-1} (AB - \sqrt{1-A^2}\sqrt{1-B^2})$

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2\sqrt{7}}$

Vedclass · Answer

(A) Given $\cos ^{-1} 2x + \cos ^{-1} 3x = \frac{\pi}{3}$.Using the formula $\cos ^{-1} A + \cos ^{-1} B = \cos ^{-1} (AB - \sqrt{1-A^2}\sqrt{1-B^2})$, we get:$\cos ^{-1} (2x \cdot 3x - \sqrt{1-(2x)^2}\sqrt{1-(3x)^2}) = \frac{\pi}{3}$$6x^2 - \sqrt{1-4x^2}\sqrt{1-9x^2} = \cos \frac{\pi}{3} = \frac{1}{2}$$6x^2 - \frac{1}{2} = \sqrt{(1-4x^2)(1-9x^2)}$Squaring both sides:$(6x^2 - \frac{1}{2})^2 = (1-4x^2)(1-9x^2)$$36x^4 - 6x^2 + \frac{1}{4} = 1 - 13x^2 + 36x^4$$-6x^2 + 13x^2 = 1 - \frac{1}{4}$$7x^2 = \frac{3}{4}$$x^2 = \frac{3}{28}$$x = \sqrt{\frac{3}{28}} = \frac{\sqrt{3}}{2\sqrt{7}}$

If $\cos ^{-1} 2x + \cos ^{-1} 3x = \frac{\pi}{3}$, then $x =$

Similar Questions

If $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2}$,then $x^{100}+y^{100}+z^{100}=$

The value of $x$ which satisfies $\sin \left(\cot ^{-1} x\right)=\cos \left(\tan ^{-1}(1+x)\right)$ is

If $0 < x < 1$,then $\cot ^{-1}\left( \frac{2x^2 - 1}{2x\sqrt{1 - x^2}} \right)$ is equal to

If $\sin^{-1}(x - 2) + \cos^{-1}(x) + \tan^{-1}(x + 2) + \cot^{-1}(x + 4) = \sec^{-1}(\sqrt{k}) - \frac{\pi}{2}$, then $\cos(2 \csc^{-1}\sqrt{k - 1}) = \dots$

$\cos ^{-1}\left(\frac{-1}{2}\right)-2 \sin ^{-1}\left(\frac{1}{2}\right)+3 \cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)-4 \tan ^{-1}(-1)$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module