Solve a trigonometric equation

[anemone]

Solve the equation

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

 

[Prove It]

Solve the equation

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

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

 

[mathworker]

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

 

[MarkFL]

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

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. (Wink)

 

[kaliprasad]

Solve the equation

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

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

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