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

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

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2sin^2(x)+7cos(x)=5

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

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Solution steps

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

Subtract

5

from both sides

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

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

cos (x) = 3 :

No Solution

Combine all the solutions

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

2 sin^{2} (θ) - 3 sin (θ) + 1 = 0, 0 \leq θ < 2 π

2 sin^{2} (x) = 3 sin (x)

4 sin^{2} (x) - 3 = 0, 0 \leq x \leq 2 π

cos (x) - cot (x) = 0

6 sin (θ) + 7 = 0

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