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#