Answer to Question #93165 in Trigonometry for Amey

Question #93165
IfSin⁴x-cos^7x=1 find general solution
1
Expert's answer
2019-08-23T09:45:04-0400

"\\sin ^{4} x-\\cos ^{7} x=1 \\\\\n\\frac{3-4 \\cos 2 x+\\cos 4 x}{8}-\\cos ^{7} x=1 \\\\\n3-4\\left(2 \\cos ^{2} x-1\\right)+\\left(2 \\cos ^{2} 2 x-1\\right)-8 \\cos ^{7} x=8 \\\\\n3-8 \\cos ^{2} x+4+2\\left(2 \\cos ^{2} x-1\\right)^{2}-1-8 \\cos ^{7} x=8 \\\\\n-8 \\cos ^{2} x-8 \\cos ^{7} x+8 \\cos ^{4} x-8 \\cos ^{2} x+2+6=8 \\\\\n\\cos ^{4} x-\\cos ^{7} x-2 \\cos ^{2} x=0 \\\\\n\\cos ^{2} x\\left(\\cos ^2{x}-\\cos ^{5} x-2\\right)=0 \\\\\n\\cos x=0 \\\\\nx=\\frac{\\pi}{2}+\\pi k, k \\in z \\\\\n\\cos ^{2} x-\\cos ^{5} x-2=0 \\\\\n\\begin{array}{l}{\\cos x=-1} \\\\ {x=\\pi+2 \\pi k, k \\in Z}\\end{array} \\\\\n\n\\begin{array}{c}{\\text { Answer: } \\frac{\\pi}{2}+\\pi k, k \\in Z} \\ \\ \\ \\ ; \\ \\ \\ \\ {\\pi+2 \\pi k, k \\in Z}\\end{array}"


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