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

The general solution for 4sin^2(x/2)+8cos(x/2)-7=0 is x=(2pi)/3+4pin,x=(10pi)/3+4pin

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4sin^2(x/2)+8cos(x/2)-7=0

4 sin^{2} (\frac{x}{2}) + 8 cos (\frac{x}{2}) - 7 = 0

x = \frac{2 π}{3} + 4 πn, x = \frac{10 π}{3} + 4 πn

Solution steps

4 sin^{2} (\frac{x}{2}) + 8 cos (\frac{x}{2}) - 7 = 0

Rewrite using trig identities

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

Solve by substitution

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

cos (\frac{x}{2}) = \frac{1}{2} : x = \frac{2 π}{3} + 4 πn, x = \frac{10 π}{3} + 4 πn

cos (\frac{x}{2}) = \frac{3}{2} :

No Solution

Combine all the solutions

x = \frac{2 π}{3} + 4 πn, x = \frac{10 π}{3} + 4 πn

sin (x) + cos (x) = 1, - 36 0^{\circ} \leq x \leq 36 0^{\circ}

5 csc (θ) + 6 = 0

sin (x^{2}) = 1

cos (2 θ) - 3 cos (θ) = 0, 0 \leq θ < 2 π

2 cos^{2} (θ) - 7 cos (θ) - 4 = 0

What is the general solution for 4sin^2(x/2)+8cos(x/2)-7=0 ?
The general solution for 4sin^2(x/2)+8cos(x/2)-7=0 is x=(2pi)/3+4pin,x=(10pi)/3+4pin