Nice question by combining logarithm and trigonometry

Algebra Level 3

If log ⁡ 24 sin ⁡ x ( 24 cos ⁡ x ) = 3 2 \log_{24\sin x}(24\cos x)=\dfrac{3}{2} , find the value of cot ⁡ 2 x \cot^2x for x ∈ ( 0 , π 2 ) x\in \left(0,\dfrac{\pi}{2}\right) .


The answer is 8.

Jitendra Kumar by Jitendra Kumar , 28, India

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1 solution

log ⁡ 24 sin ⁡ x ( 24 cos ⁡ x ) = 3 2 ( 24 sin ⁡ x ) 3 2 = 24 cos ⁡ x Squaring both sides. 2 4 3 sin ⁡ 3 x = 2 4 2 cos ⁡ 2 x 24 sin ⁡ 3 x = cos ⁡ 2 x = 1 − sin ⁡ 2 x 24 sin ⁡ 3 x + sin ⁡ 2 x − 1 = 0 ⟹ sin ⁡ x = 1 3 \begin{aligned} \log_{24\sin x}(24\cos x) & = \frac 32 \\ (24\sin x)^\frac 32 & = 24 \cos x & \small \color{#3D99F6} \text{Squaring both sides.} \\ 24^3 \sin^3 x & = 24^2 \cos^2 x \\ 24 \sin^3 x & = \cos^2 x = 1 - \sin^2 x \\ 24 \sin^3 x + \sin^2 x - 1 & = 0 \\ \implies \sin x & = \frac 13 \end{aligned}

Therefore, cot ⁡ 2 x = cos ⁡ 2 x sin ⁡ 2 x = 1 − sin ⁡ 2 x sin ⁡ 2 x = 1 sin ⁡ 2 x − 1 = 1 ( 1 3 ) 2 − 1 = 8 \cot^2 x = \dfrac {\cos^2 x}{\sin^2 x} = \dfrac {1-\sin^2 x}{\sin^2 x} = \dfrac 1{\sin^2 x} - 1 = \dfrac 1{\left(\frac 13\right)^2}-1 = \boxed{8} .

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how did u proceed to 2nd step? \

rakshith lokesh - 3 years, 3 months ago

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log ⁡ x y = a ⇒ y = x a \log_xy = a \Rightarrow y=x^a

Jitendra Kumar - 3 years, 3 months ago

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