How do you solve 3sin2x+4cosx4=0?

1 Answer
Feb 1, 2018

The answers are x=cos1(13)+2πk,2πk.

Explanation:

3sin2x+4cosx4=0

Use the identity sin2x+cos2x=1sin2x=1cos2x

3(1cos2x)+4cosx4=0

33cos2x+4cosx4=0

13cos2x+4cosx=0

Let u=cosx:

13u2+4u=0

3u2+4u1=0

3u24u+1=0

(3u1)(u1)=0

u=13,1

Now plug cosx back in for u:

cosx=13,cosx=1

x=cos1(13)+2πk,2πk

where k is any integer. For example, x could be 0, 2π, 4π, etc.