Solve $[\sin(\frac{\pi}{5})]^{x}+[\cos(\frac{\pi}{5})]^x$=1

Question

Solve $[\sin(\frac{\pi}{5})]^{x}+[\cos(\frac{\pi}{5})]^x=1$

Here is another problem that I don't know how to solve. I know that $\sin^2(x)+\cos^2(x)=1$. How do I prove that here $x=2$?. Isn't this just a particular case of proving $\sin^2(kx)+\cos^2(kx)=1$ with $k$ being an integer?

Answer 1

Because both $sin\left(\frac\pi5\right)$ and $cos\left(\frac\pi5\right)$ are fractions, each will become smaller as you raise them to successively higher powers. At $x=2$ they already add to exactly $1$. Adding two smaller numbers will never add up to $1$.

Answer 2

http://www.wolframalpha.com/input/?i=sin+(pi%2F5)%5Ex+%2B+cos+(pi%2F5)%5Ex+%3D+1 This might help by equating LHS to RHS.

Here $(\frac {5}{8}-\frac {\sqrt 5}{8})^{\frac x2}+(\frac 14(1+\sqrt5))^x=1$. Simplify and solve equation you will get x=2. Also plotting graph will give the rational solution.

Answer 3

Hints:

1. Rewrite the equation as $y = (\sin \frac {\pi}{5})^x + (\cos \frac {\pi}{5})^x - 1$ and graph it. You've seen that $x=2$ is a solution; is there another one (or more)?
2. When you look at the graph, what do you notice after $x > 2$? What about $x<0?$
3. If you attempt to solve the equation using logarithms, what do you notice?

Answer 4

$$\sin(\pi/5)^x+\cos(\pi/5)^x = 1$$.
since we know that $\cos(x) = \sqrt(1-\sin(x)^2)$ w'll use that identity.
$$\sin(\pi/5)^x -1 = -1*\cos(\pi/5)^x$$ $$\sin(\pi/5)^x - 1 = ( -1*\sqrt(1-\sin(\pi/5)^2 )^x$$ $$(\sin(\pi/5)^x - 1)^2 = ( 1-\sin(\pi/5)^2 )^x$$.
by simple comparing the two sides of the equation one can see that $x = 2$ is a solution.