What is the general solution for 2sin^2(x)-7cos(x)-5=0 ?

The general solution for 2sin^2(x)-7cos(x)-5=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin

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2sin^2(x)-7cos(x)-5=0

2 sin^{2} (x) - 7 cos (x) - 5 = 0

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

Solution steps

2 sin^{2} (x) - 7 cos (x) - 5 = 0

Rewrite using trig identities

- 3 - 2 cos^{2} (x) - 7 cos (x) = 0

Solve by substitution

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

cos (x) = - 3 :

No Solution

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

Combine all the solutions

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

cos (x) = sin (0)

1 = \frac{1}{0.5} sin (x + 3 0^{\circ})

- tan^{2} (θ) = - 1 + sec (θ)

x, y = sin (x^{c o s (y)})

3 sec (θ) + 3 = 0

What is the general solution for 2sin^2(x)-7cos(x)-5=0 ?
The general solution for 2sin^2(x)-7cos(x)-5=0 is x=(2pi)/3+2pin,x=(4pi)/3+2pin