If 32 ^8 θ = 2 ² α - 3 α and 3 2 θ = 1 then the general value of α for n Z is

Correct Answer is : 2nπ±2π32 n π ± 2π/32nπ±32π

Correct Answer is : 2nπ±2π32 n π ± 2π/32nπ±32π

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NIMCET 2021 Question Paper with solutions

Question : 28 of 120

Marks: +1, -0

32 \tan^8 \theta = 2 \cos^2 \alpha - 3 \cos \alpha

3 \cos 2 \theta = 1

\alpha

n \in Z

Solution:

32 \tan^8 \theta = 2 \cos^2 \alpha - 3 \cos \alpha

and

3 \cos 2 \theta = 1

then

\alpha =

3 \cos 2 \theta = 1

\cos 2 \theta = \frac{1}{3}

\cos 2 \theta = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}

\frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} = \frac{1}{3}

3 - 3 \tan^2 \theta = 1 + \tan^2 \theta

2 = 4 \tan^2 \theta

\tan^2 \theta = \frac{1}{2}

\tan^8 \theta = \frac{1}{16}

32 \tan^8 \theta = 2 \cos^2 \alpha - 3 \cos \alpha

32 \times \frac{1}{6} = 2 \cos^2 \alpha - 3 \cos \alpha

2 = 2 \cos^2 \alpha - 3 \cos \alpha

2 \cos^2 \alpha - 3 \cos \alpha - 2 = 0

\cos \alpha = \frac{3 \pm \sqrt{9+16}}{2 \times 2}

\cos \alpha = \frac{3 \pm 5}{4}

\cos \alpha = 2, \frac{-1}{2}

-1 \leq \cos \alpha \leq 1

\cos \alpha = -\frac{1}{2}

\alpha = 2 n \pi \pm \frac{2\pi}{3}

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