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

The general solution for 4sin^2(x)+4cos(x)-5=0,(0,2pi) is x= pi/3 ,x=(5pi)/3

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

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

x = \frac{π}{3}, x = \frac{5 π}{3}

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

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

cos (x) = \frac{1}{2}, (0, 2 π) : x = \frac{π}{3}, x = \frac{5 π}{3}

Combine all the solutions

x = \frac{π}{3}, x = \frac{5 π}{3}

a, E = 25 (sin (a) - cos (a))

391 = 78 \cdot 21.5 \cdot cos (x)

5 tan (θ) - 9 = - 6 tan (θ) + 8

81 (cos^{2} (x)) = 6

(cos (x) - 2) (cos (x) + 1) = 0

What is the general solution for 4sin^2(x)+4cos(x)-5=0,(0,2pi) ?
The general solution for 4sin^2(x)+4cos(x)-5=0,(0,2pi) is x= pi/3 ,x=(5pi)/3