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

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

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

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

x = \frac{π}{6}, x = \frac{5 π}{6}

Solution steps

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

Subtract

5 - 4 sin (x)

from both sides

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

Rewrite using trig identities

- 1 + 4 sin (x) - 4 sin^{2} (x) = 0

Solve by substitution

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

sin (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{6}, x = \frac{5 π}{6}

Combine all the solutions

x = \frac{π}{6}, x = \frac{5 π}{6}

0 = 6 sin (x)

sec (2 x) = - 1

tan (2 x) = \frac{80}{\frac{30 + 20}{2}}

tan^{2} (x) + 1 = 4 tan (x)

(sin (x) - 1) (cos (x) - \frac{1}{2}) = 0

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