What is the general solution for 4sin^2(θ/2)+8cos(θ/2)-7=0,0<= θ<= 360 ?

The general solution for 4sin^2(θ/2)+8cos(θ/2)-7=0,0<= θ<= 360 is θ=120

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4sin^2(θ/2)+8cos(θ/2)-7=0,0<= θ<= 360

4 sin^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2}) - 7 = 0, 0 \leq θ \leq 36 0^{\circ}

θ = 12 0^{\circ}

Solution steps

4 sin^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2}) - 7 = 0, 0 \leq θ \leq 36 0^{\circ}

Rewrite using trig identities

- 3 - 4 cos^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2}) = 0

Solve by substitution

cos (\frac{θ}{2}) = \frac{1}{2}, cos (\frac{θ}{2}) = \frac{3}{2}

cos (\frac{θ}{2}) = \frac{1}{2}, 0 \leq θ \leq 36 0^{\circ} : θ = 12 0^{\circ}

cos (\frac{θ}{2}) = \frac{3}{2}, 0 \leq θ \leq 36 0^{\circ} :

No Solution

Combine all the solutions

θ = 12 0^{\circ}

- cos (θ) + 1 = 2 cos (θ) + 3

cot^{2} (x) + cot (x) = 0

3 cos (2 θ) + 9 cos (θ) - 4 = - 2 cos (θ)

16 cos^{2} (x) - 12 = 0

cos (2 θ) - cos^{2} (θ) = 0

What is the general solution for 4sin^2(θ/2)+8cos(θ/2)-7=0,0<= θ<= 360 ?
The general solution for 4sin^2(θ/2)+8cos(θ/2)-7=0,0<= θ<= 360 is θ=120