The number of solutions of equation 6 cos 2θ + 2 cos2(θ/2) + 2 sin2 θ = 0, -π

Question

The number of solutions of equation 6 cos 2θ + 2 cos2(θ/2) + 2 sin2 θ = 0, -π < θ < π is (A) 3 (B) 4 (C) 5 (D) 6

Tardigrade Admin · Accepted Answer

6 cos 2θ + 2 cos2(θ/2) + 2 sin2θ = 0 or 12 cos2θ - 6 + 1 + cos θ + 2 - 2 cos2θ = 0 or 10 cos2 θ + cos θ - 3 = 0 or (5 cos θ + 3) (2 cos θ - 1) = 0 or cos θ = -(3/5), (1/2) ⇒ θ = (π/3), π - cos-1 ((3/5)), - (π/3), - π + cos-1 ((3/5))

Q. The number of solutions of equation $6 cos 2 θ + 2 co s^{2} (θ /2) + 2 s i n^{2} θ = 0, - π < θ < π$ is

$3$

$4$

$5$

$6$

Q. The number of solutions of equation 6cos2θ+2cos2(θ/2)+2sin2θ=0,−π<θ<π is

3

4

5

6

6cos2θ+2cos2(θ/2)+2sin2θ=0 or 12cos2θ−6+1+cosθ+2−2cos2θ=0 or 10cos2θ+cosθ−3=0 or (5cosθ+3)(2cosθ−1)=0 or cosθ=−53​,21​ ⇒θ=3π​,π−cos−1(53​),−3π​,−π+cos−1(53​)

Q. The number of solutions of equation $6 cos 2 θ + 2 co s^{2} (θ /2) + 2 s i n^{2} θ = 0, - π < θ < π$ is

$3$

$4$

$5$

$6$