Question

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

Answer 1

`cosx =(7pm1)/12`

Explanation:

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

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`