Answer to Question #147769 in Trigonometry for Moustafa Elemam

Question #147769
(21 - 16〖sin〗^2 x+8 cosx)/(16 〖cos〗^2 x-29-8 √15 sin⁡x ) = 2

find x
1
Expert's answer
2020-12-01T06:23:09-0500
"\\dfrac{21-16\\sin^2x+8\\cos x}{16\\cos^2x-29-8\\sqrt{15}\\sin x}=2"

The Pythagorean Identity


"\\sin^2x+\\cos^2x=1"


Then


"\\dfrac{5+16(1-\\sin^2x)+8\\cos x}{16(\\cos^2x-1)-13-8\\sqrt{15}\\sin x}=2"

"\\dfrac{16\\cos^2x+8\\cos x+5}{-16\\sin^2x-8\\sqrt{15}\\sin x-13}=2"

Let "u=\\sin x, v=\\cos x." Then


"\\begin{cases}\n u^2+v^2=1 \\\\\n 16v^2+8v+5=-32u^2-16\\sqrt{15}u -26\n\\end{cases}"




"\\begin{cases}\n u^2+v^2=1 \\\\\n (16v^2+8v+1)+2(16u^2+8\\sqrt{15}u +15)=0\n\\end{cases}"


"\\begin{cases}\n u^2+v^2=1 \\\\\n (4v+1)^2+2(4u+\\sqrt{15})^2=0\n\\end{cases}"

Then

"\\begin{cases}\n u^2+v^2=1 \\\\\n 4v+1=0 \\\\\n4u+\\sqrt{15}=0\n\\end{cases}"

Check


"-16\\sin^2x-8\\sqrt{15}\\sin x-13\\not=0"

"-16u^2-8\\sqrt{15}u-13\\not=0"

"u=-\\dfrac{\\sqrt{15}}{4}:"


"-16(-\\dfrac{\\sqrt{15}}{4})^2-8\\sqrt{15}(-\\dfrac{\\sqrt{15}}{4})-13"

"=-15+30-13=2\\not=0"

Hence

"\\cos x=-\\dfrac{1}{4}"

"x=\\pm(\\pi-\\arccos(\\dfrac{1}{4}))+2\\pi n, n\\in\\Z"


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