NIMCET 2021 Question Paper with solutions

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Question : 28 of 120

Marks: +1, -0

If

32 \tan ^{8} \θ=2 \cos ^{2} \α-3 \cos \α

and

3 \cos 2 \θ=1

then the general value of

\α

for

{n} \∈{Z}

is

${n} \π\±{\π}/{3}$
$2 n \π\±{2 \π}/{3}$
$2 n \π\±{\π}/{3}$
$n \π\±{2 \π}/{3}$

Solution:

32 \tan ^{8} \θ=2 \cos ^{2} \α-3 \cos \α

and

3 \cos 2 \θ=1

then

\α=

3 \cos 2 \θ=1

\cos 2 \θ={1}/{3}

\cos 2 \θ={1-\tan ^{2} \θ}/{1+\tan ^{2} \θ}

{1-\tan ^{2} \θ}/{1+\tan ^{2} \θ}={1}/{3}

3-3 \tan ^{2} \θ=1+\tan ^{2} \θ

2=4 \tan ^{2} \θ

\tan ^{2} \θ={1}/{2}

\tan ^{8} \θ={1}/{16}

32 \tan ^{8} \θ=2 \cos ^{2} \α-3 \cos \α

32 \× 1 \/ 6=2 \cos ^{2} \α-3 \cos \α

2=2 \cos ^{2} \α-3 \cos \α

2 \cos ^{2} \α-3 \cos \α-2=0

\cos \α={3 \±\√{9+16}}/{2 \×2}

\cos \α={3 \± 5}/{4}

\cos \α=2, {-1}/{2}

as

-1 \≤ \cos \α \≤ 1

So

\cos \α=-{1}/{2}

\α=2 {n} \π\±{2 \π}/{3}

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