Answer to Question #93165 in Trigonometry for Amey

Question

IfSin⁴x-cos^7x=1 find general solution

Answer 1

$\sin ^{4} x-\cos ^{7} x=1 \\ \frac{3-4 \cos 2 x+\cos 4 x}{8}-\cos ^{7} x=1 \\ 3-4\left(2 \cos ^{2} x-1\right)+\left(2 \cos ^{2} 2 x-1\right)-8 \cos ^{7} x=8 \\ 3-8 \cos ^{2} x+4+2\left(2 \cos ^{2} x-1\right)^{2}-1-8 \cos ^{7} x=8 \\ -8 \cos ^{2} x-8 \cos ^{7} x+8 \cos ^{4} x-8 \cos ^{2} x+2+6=8 \\ \cos ^{4} x-\cos ^{7} x-2 \cos ^{2} x=0 \\ \cos ^{2} x\left(\cos ^2{x}-\cos ^{5} x-2\right)=0 \\ \cos x=0 \\ x=\frac{\pi}{2}+\pi k, k \in z \\ \cos ^{2} x-\cos ^{5} x-2=0 \\ \begin{array}{l}{\cos x=-1} \\ {x=\pi+2 \pi k, k \in Z}\end{array} \\ \begin{array}{c}{\text { Answer: } \frac{\pi}{2}+\pi k, k \in Z} \ \ \ \ ; \ \ \ \ {\pi+2 \pi k, k \in Z}\end{array}$