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+3 votes
43.8k views
in JEE by (130k points)

(1)-3/5
(2) 1/3
(3) 2/9
(4) -7/9

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2 Answers

+5 votes
by (14.7k points)

Answer is -7/9

5 (tan2x - cos2x) = 2cos2x +9 

5 ( (sec2x - 1) - cos2x) = 2(2cos2x - 1) +9 

5 (1/cos2x - 1- cos2x) = 4cos2x - 2 + 9 

5 (1 - cos2x - cos4x ) = 4cos4x + 7cos2

You solve this we get = 9cos4x+12cos2x-5 

Therefore let cos4x =y2 and cos2x = y. 

0=9y2 + 12y - 5

You will get the answer by solving for y

0 votes
by (38.1k points)

5(tan2 x - cos2 x) = 2 cos 2x + 9

= 5(sec2 x - 1 - cos2 x) = 2(2cos2 x - 1) + 9 (\(\because\) cos2x = 2cos2 x - 1)

= 5 \((\cfrac{1}{cos^2x}- 1 -cos^2x)\) = 4 cos2 x -2 + 9  (\(\because\) sec x = \(\cfrac{1}{cos x}\))

= 5(1 - cos2 x - cos4x) = 4 cos4x +7 cos2 x ( multiplying both sides by cos2x)

= - 5 cos4 x - 5 cos2 x + 5 = 4 cos4x + 7 cos2

= 9 cos4x + 12 cos2x - 5 = 0 (by transposing )

which is a quadric equation in cos2 x.

= 9 cos4x + 15 cos2 x - 3 cos2 x - 5 = 

= 3 cos2 x(3 cos2 x + 5) - 1(3 cos2 x + 5) = 0

= (3 cos2 x + 5) (3 cos2 x - 1) = 0

= 3cos2 x + 5 = 0 or 3 cos2 x - 1 = 0

= cos2 x = \(\frac{-5}3\) or cos2 x = \(\frac{1}3\)

= cos2x = \(\frac{1}3\) = (\(\because\) cos2 x ≥ 0)

= cos2 x ≠ \(\frac{-5}3\) )

Now cos2 x = 2 cos2 x - 1

\(\frac{2}3\) - 1 = \(\frac{-1}3\)

\(\therefore\) cos4 x = 2cos2(2x) - 1 (\(\because\) cos 2\(\theta\) = 2cos2\(\theta\) - 1)

= 2 \((\frac{-1}3)^2\) - 1

\(\frac{2}9\) - 1

\(\frac{2-9}9\) = \(\frac{-7}9\)

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