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

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

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2sin^2(x)+7cos(x)+2=0

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

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

Solution steps

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

Rewrite using trig identities

4 - 2 cos^{2} (x) + 7 cos (x) = 0

Solve by substitution

cos (x) = - \frac{1}{2}, cos (x) = 4

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

cos (x) = 4 :

No Solution

Combine all the solutions

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

1000 cos (4 5^{\circ}) = 520 cos (3 5^{\circ}) + 1043.937687 cos (θ)

4 sin (\frac{x}{2}) = 4 cos (\frac{x}{2})

tan^{2} (θ) + 8 tan (θ) + 9 = 0

2 tan^{2} (x) - 1 = 3 tan (x)

8 csc^{2} (x) - 2 cot (x) = 9

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