How do you answer the following question, I know it's a trigonometric equation but does anyone know how to turn it into a form of cos/sin x = number? 6sin^2(x)+7cos(x)=8 Thank you

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Answer 1 · 2017-02-15T12:51:28

$cosx =(7pm1)/12$

Using the identity

$cos^2x+sin^2x=1$ and substituting we get

$6(1-cos^2x)+7cosx-8=0$ or

$6cos^2x-7cosx+2=0$ . Now solving for $cosx$

$cosx=(7pmsqrt(49-48))/12=(7pm1)/12$

Answer 2 · 2017-02-15T16:14:26

We are to turn the given equation into a form of $cosx/sinx="number"$

Let me try to have a solution.

Given equation is

$6sin^2x+7cosx=8$

Dividing both sides by sin^2x we get

$6+(7cosx)/sin^2x=8/sin^2x$

$=>6+7cotxcscx=8csc^2x$

$=>6+7cotxcscx=8(1+cot^2x)$

$=>6+7cotxcscx=8+8cot^2x$

$=>8cot^2x+8-6=7cotxcscx$

$=>8cot^2x+2=7cotxcscx$

$=>(8cot^2x+2)^2=7^2cot^2xcsc^2x$

$=>(8cot^2x+2)^2=7^2cot^2x(1+cot^2x)$

$=>64cot^4x+32cot^2x+4-49cot^4x-49cot^2x=0$

$=>15cot^4x-17cot^2x+4=0$

$=>15cot^4x-12cot^2x-5cot^2x+4=0$

$=>3cot^2x(5cot^2x-4)-1(5cot^2x-4)=0$

$=>(3cot^2x-1)(5cot^2x-4)=0$

When $3cot^2x-1=0$

$=>cotx=pm1/sqrt3$

$=>cosx/sinx=pm1/sqrt3$

Again when

$5cot^2x-4=0$

$=>cotx=pm2/sqrt5$

$=>cosx/sinx=pm2/sqrt5$