solve 1+sinθ=2cos2θ on the interval 0≤θ<2π.

Answer 1

Hi,

1+sinθ=2cos2θ

first, you need to change cos2θ to the sinθ

you Know: cos( 2a ) = 1 - 2 sin² (a) then replace to the equation

1+sina=2 ( 1 - 2 sin² (a) ) please let me change theta to a because of my laptop

Now simplify: 4sin²(a) + sin (a ) -1 = 0 solve this equation by quadratic formula

you will have 2 different answers such as sin (a) = (-1 + √17)/8 and sin (a) = (-1 - √17)/8

use calculator and find inverse of sin

sin (a) = (-1 + √17)/8 ------> a = 22.97 , 157.02

sin (a) = (-1 - √17)/8 ------> a = 320.18 , 219.82

Hope it was useful and clear, please let me know if you have any question.

Minoo

Answer 2

210 or 330 degrees or

7pi/6 or 11pi/6 radians

1 + sinx = 2cos2x

1+sinx = 2(1-2sin^2x) = 2-4sin^2x

1-2 = -4sin^2x/sinx

-1 = 4sin^2x/sinx = 4sinx

sinx = -1/4

x = sin^-1(-1/4)

x = 210 or 330 degrees = 7pi/6 or 11pi/6 radians

cos2x = cos^2x - sin^2x

cos^2x + sin^2x = 1

cos^2x = 1 - sin^2x

cos2x = 1-sin^2x - sin^2x = 1 - 2sin^2x