Solve a trigonometric equation

anemone

MHB POTW Director
Staff member
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

Well-known member
MHB Math Helper
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

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

Last edited:

MarkFL

Staff member
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.

Well-known member
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