solve the equation on the interval 0≤θ<2π

Question

1. -1 -8cos(-30+ pi/6) = 3
2. 1 +4sin^2 θ = -4sin θ
3. 2cos^2 θ = -cos2 θ +1

Answer 1

1) -1-8cos(-3x+pi/6)= 3

cos(-3x+pi/6)=-1/2

-3x+pi/6= 2pi/3, 4pi/3, 8pi/3, 10pi/3, 4pi, 14pi/3, 16pi/3, 18pi/3

-3x= pi/2, 7pi/6, 15pi/6, 19pi/6, 23pi/6, 27pi/6, 31pi/6, 35pi/6

x = -pi/6, -7pi/18, -5pi/6, -19pi/18, -23pi/18, -27pi/18, -31pi/18, -35pi/18

x= pi/18, 5pi/18, pi/2, 13pi/18, 17pi/18, 7pi/6, 29pi/18, 11pi/6

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

let y=sinx

1+4y^2 =-4y

4y^2 +4y+1=0

(2y+1)^2 = 0

2y+1=0

2y=-1

y=-1/2

sinx=-1/2

x= 7pi/6, 11pi/6

3) 2cos^2(x)= -cos2x+1

2cos^2(x)-1 =-cos2x

cos2x =-cos2x

2cos2x=0

cos2x =0

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

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