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5(tan^2x-cos^2x)=2cos2x+9 find cos4x
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5(tan^2x-cos^2x)=2cos2x+9 find cos4x
5(sec^2 x – 1 – cos^2 x) = 2(2cos^2 x – 1) + 9Let cos^2 x = t5(1/t - 1 - t) = 2(2t - 1) + 95(1 – t – t^2) = 4t^2 + 7t9t^2 + 12t – 5 = 0t = 1/3, -5/3cos^2 x = 1/3cos 2x = 2/3 - 1 = -1/3cos 4x = -7/9
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5(tan^2x-cos^2x)=2cos2x+9 find cos4x
5 (tan^2x - cos^2x) = 2cos2x +95 ( (sec^2x - 1) - cos^2x) = 2(2cos^2x - 1) +95 (1/cos^2x - 1- cos^2x) = 4cos^2x - 2 + 95 (1 - cos^2x - cos^4x ) = 4cos^4x + 7cos^2xYou solve this we get = 9cos^4x+12cos^2x-5Therefore let cos^4x =y^2 and cos^2x = y.0=9y^2 + 12y - 5You solve this we get answer.
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5(tan^2x-cos^2x)=2cos2x+9 find cos4x
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