solve the equation giving the exact solutions which lie in 0

solve the equation, giving the exact solutions which lie in [0, 2)

A) sin(2x) = tan(x)

B) sin(x) + 3 cos(x) = 1

C) cos(3x) = cos(5x)

a) sin2x = tanx

2sinxcosx = sinx/cosx

sinx gets cancelled out

2cosx = 1/cosx

2cos^2x =1

2cos^2x -1 =0

(sqrt2cosx+1)(sqrt2cosx -1) =0

cosx = -1/sqrt2

x = pi -pi/4 , pi +pi/4

x = 3pi/4 , 5pi/4

cosx = 1/sqrt2

x = pi/4 , 2pi -pi/4

= pi/4 , 7pi/4

x = pi/4 , 3pi/4 , 5pi/4 , 7pi/4

b) sinx + sqrt3cosx =1

sinx = 1- sqrt3cosx

square both sides:

sin^2x = ( 1- sqrt3cosx)^2

(1- cos^2x ) - ( 1- sqrt3cosx)^2 =0

-4cos^2x + 2cosx*sqrt3=0

solve the quadratic:

cosx =0 ; cosx = sqrt3/2

x =pi/2 , 11pi/6

c) cos(3x) = cos(5x)

cos(3x ) - cos(5x) =0

use formula: cosA - cosB = -2sin(A+B)/2sin(A -B)/2

cos(3x ) - cos(5x) = 2sin(4x)sinx

2sin4xsinx =0

sin4x =0

x= 0 ,pi/4, pi/4 , 3pi/4, pi , 5pi/4 , 3pi/2 , 7pi/4 , 2pi

sinx =0

x = 0 , pi , 2pi

combining the two results : x= 0 ,pi/4, pi/4 , 3pi/4, pi , 5pi/4 , 3pi/2 , 7pi/4 , 2pi