Question

Find at least two x that make this equation true:

sin [x + {(π^3 + 2√ π^6)/(π^2 + π^2)} + π^π^0 ] = cos [x + {((−1)^16 / 2)} − {log2 ( √ 8) / 3 }]

Answer 1

Answer:

0, π/2

Step-by-step explanation:

$\frac{\pi^3+2\sqrt{\pi^6}}{\pi^2+\pi^2} +\pi^{\pi^0}$

$=\frac{\pi^3+2\pi^3}{2\pi^2} +\pi^{1}$

$=\frac{3\pi^3}{2\pi^2} +\pi$

$=\frac{3\pi}{2} +\pi$

$=\frac{5\pi}{2}$

$=2\pi +\frac{\pi}{2}$

Again

$\frac{(-1)^{16}}{2}-\frac{\log_2\sqrt{8}}{3}$

$=\frac{1}{2}-\frac{\log_2 2^{3/2}}{3}$

$=\frac{1}{2}-\frac{(3/2)\log_2 2}{3}$

$=\frac{1}{2}-\frac{1}{2}$

$=0$

Therefore,

$\sin(x+\frac{\pi^3+2\sqrt{\pi^6}}{\pi^2+\pi^2} +\pi^{\pi^0})=\cos(\frac{(-1)^{16}}{2}-\frac{\log_2\sqrt{8}}{3})$

$\implies \sin(x+2\pi +\frac{\pi}{2})=\cos(x+0)$

$\implies \sin(2\pi +\frac{\pi}{2}+x)=\cos(x)$

$\implies \sin(\frac{\pi}{2}+x)=\cos(x)$

Which is true for all x ∈ R

Therefore, two values that satisfy the equation can be $0, \frac{\pi}{2}$

Hope this helps.