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- Feb 14, 2012

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\(\displaystyle \cos^2 \left(\frac{\pi}{4}(\sin x+\sqrt{2}\cos^2 x)\right)-\tan^2 \left(x+\frac{\pi}{4}\tan^2 x\right)=1\)

- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

- 3,963

\(\displaystyle \cos^2 \left(\frac{\pi}{4}(\sin x+\sqrt{2}\cos^2 x)\right)-\tan^2 \left(x+\frac{\pi}{4}\tan^2 x\right)=1\)

I would probably start by moving the tangent function to the other side and applying the Pythagorean Identity to convert everything to cosines...

\(\displaystyle \cos^2 \left(\frac{\pi}{4}(\sin x+\sqrt{2}\cos^2 x)\right)-\tan^2 \left(x+\frac{\pi}{4}\tan^2 x\right)=1\)

- May 31, 2013

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yeah there by it will become product of 2 cosine's =1 that means both equal to 1

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People posting problems in this sub-forum are presumed to already have worked out the solution in full and are posting the problem as a challenge to others to solve rather than asking for help.I would probably start by moving the tangent function to the other side and applying the Pythagorean Identity to convert everything to cosines...

- Mar 31, 2013

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as cos^2 t < = 1 and tan ^2 a >= 0 so the 1st term is 1 and 2nd term is zero.

\(\displaystyle \cos^2 \left(\frac{\pi}{4}(\sin x+\sqrt{2}\cos^2 x)\right)-\tan^2 \left(x+\frac{\pi}{4}\tan^2 x\right)=1\)

So cos^2(π/4(sinx+√2 cos2x)) = 1 and tan^2(x+π/4tan^2x) = 0

So (x+π/4tan^2 x) = n π , x = 0 is a solution and other solutions are

= n π – π/4(this I found by guessing tan ^2x =1 and not rigorously)

cos^2(π/4(sin x+√2cos^2x)) = 1 when x = 2 n π – π/4

hence this is the solution