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

The general solution for 2cos^2(x)+7sin(x)-5=0 is x= pi/6+2pin,x=(5pi)/6+2pin

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

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

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

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

sin (x) = \frac{1}{2} : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

sin (x) = 3 :

No Solution

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

\frac{( cos ( x ) cot ( x ) )}{1 - sin ( x )} = 3

sin (2 x) - (4) \cdot cos (x) = 0

cos (θ) = \frac{2}{7}

8 tan (2 x) - 16 cos (x) = 0

x, sin (2 x) = sin (2 y)

What is the general solution for 2cos^2(x)+7sin(x)-5=0 ?
The general solution for 2cos^2(x)+7sin(x)-5=0 is x= pi/6+2pin,x=(5pi)/6+2pin