Question

Solve this trigonometric equation cos²x =3sin²x

Answer 1

Explanation:

cos²x =3sin²x subtract both sides by 3sin²x

cos²x - 3sin²x = 0 use identity cos²x+sin²x=1 => cos²x = 1-sin²x

substitute in

(1-sin²x)-3sin²x = 0 combine like terms

1-4sin²x=0 factor using difference of squares rule

(1-2sin x)(1+2sin x)=0 set each equal to 0

(1-2sin x)=0 (1+2sin x)=0

-2sinx = -1 2sinx= -1

sinx=1/2 sinx =-1/2

Think of the unit circle. When is sin x = ±1/2

at
`\pi /6, 5\pi /6, 7\pi /6, 11\pi /6`

This is from 0<x<2
`\pi`