What is the general solution for 2cos^2(x)+5sin(x)=4,0<= x<= 2pi ?

The general solution for 2cos^2(x)+5sin(x)=4,0<= x<= 2pi is x= pi/6 ,x=(5pi)/6

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2cos^2(x)+5sin(x)=4,0<= x<= 2pi

2 cos^{2} (x) + 5 sin (x) = 4, 0 \leq x \leq 2 π

x = \frac{π}{6}, x = \frac{5 π}{6}

Solution steps

2 cos^{2} (x) + 5 sin (x) = 4, 0 \leq x \leq 2 π

Subtract

4

from both sides

2 cos^{2} (x) + 5 sin (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}, 0 \leq x \leq 2 π : x = \frac{π}{6}, x = \frac{5 π}{6}

sin (x) = 2, 0 \leq x \leq 2 π :

No Solution

Combine all the solutions

x = \frac{π}{6}, x = \frac{5 π}{6}

2 - cos^{2} (x) - sin (x) - 1 = cos^{2} (x)

2 cos (2 x) + 1 = 0, 0 \leq x \leq 2 π

sec (x) + tan (x) - 1 = 0

sin (\frac{1}{2} x + \frac{π}{3}) = - \frac{1}{2}

x, tan^{2} (x) = sec (x) - 1

What is the general solution for 2cos^2(x)+5sin(x)=4,0<= x<= 2pi ?
The general solution for 2cos^2(x)+5sin(x)=4,0<= x<= 2pi is x= pi/6 ,x=(5pi)/6