What is the general solution for 5sin(x)+2cos^2(x)=4 ?

The general solution for 5sin(x)+2cos^2(x)=4 is x= pi/6+2pin,x=(5pi)/6+2pin

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5sin(x)+2cos^2(x)=4

5 sin (x) + 2 cos^{2} (x) = 4

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

Solution steps

5 sin (x) + 2 cos^{2} (x) = 4

Subtract

4

from both sides

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

Rewrite using trig identities

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

Solve by substitution

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

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

sin (x) = 2 :

No Solution

Combine all the solutions

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

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

sin^{2} (a) = 2 sin^{2} (a) + 3

cos (x) = \frac{3}{3}, 0 \leq x \leq 2 π

x, y = 2 sec (x)

- sin (2 x) - sin (x) = 0, 0 \leq x < 2 π

What is the general solution for 5sin(x)+2cos^2(x)=4 ?
The general solution for 5sin(x)+2cos^2(x)=4 is x= pi/6+2pin,x=(5pi)/6+2pin