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Solve a trigonometric equation

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
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
Jan 26, 2012
1,404
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
May 31, 2013
118
yeah there by it will become product of 2 cosine's =1 that means both equal to 1
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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

Well-known member
Mar 31, 2013
1,322
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