MHB Solve a trigonometric equation

AI Thread Summary
The equation to solve is $$\cos^2 \left(\frac{\pi}{4}(\sin x+\sqrt{2}\cos^2 x)\right)-\tan^2 \left(x+\frac{\pi}{4}\tan^2 x\right)=1$$. To approach the solution, moving the tangent function to the other side and applying the Pythagorean Identity is suggested, leading to the conclusion that both terms must equal 1 and 0 respectively. The first term equals 1 when $$\cos^2(\frac{\pi}{4}(\sin x+\sqrt{2}\cos^2 x)) = 1$$, indicating that $$x = 2n\pi - \frac{\pi}{4}$$ is a solution. The second term, $$\tan^2(x+\frac{\pi}{4}\tan^2 x) = 0$$, provides additional solutions including $$x = n\pi$$. The final solutions derived are $$x = 2n\pi - \frac{\pi}{4}$$ and $$x = n\pi$$.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Solve the equation

$$\cos^2 \left(\frac{\pi}{4}(\sin x+\sqrt{2}\cos^2 x)\right)-\tan^2 \left(x+\frac{\pi}{4}\tan^2 x\right)=1$$
 
Mathematics news on Phys.org
anemone said:
Solve the equation

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

I would probably start by moving the tangent function to the other side and applying the Pythagorean Identity to convert everything to cosines...
 
yeah there by it will become product of 2 cosine's =1 that means both equal to 1
 
Last edited:
Prove It said:
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)
 
anemone said:
Solve the equation

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

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
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top