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

The general solution for sin(2x)=5cos^2(x) is x= pi/2+2pin,x=(3pi)/2+2pin,x=1.19028…+pin

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

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

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 1.19028 \dots + πn

Solution steps

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

Subtract

5 cos^{2} (x)

from both sides

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

Rewrite using trig identities

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

Factor

- 5 cos^{2} (x) + 2 cos (x) sin (x) : cos (x) (- 5 cos (x) + 2 sin (x))

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

Solving each part separately

cos (x) = 0 or - 5 cos (x) + 2 sin (x) = 0

cos (x) = 0 : x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

- 5 cos (x) + 2 sin (x) = 0 : x = arctan (\frac{5}{2}) + πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = arctan (\frac{5}{2}) + πn

Show solutions in decimal form

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = 1.19028 \dots + πn

2 sin^{2} (x) - 3 sin (x) + 1 = 0, 0^{\circ} \leq x \leq 18 0^{\circ}

sin^{2} (θ) + cos (θ) = 2

cot (2 x) - 1 = 0

tan (2 θ) - tan (θ) = 0, 0 < θ < 2 π

- 8 cos (x) = - 6 sec (x) + 8

What is the general solution for sin(2x)=5cos^2(x) ?
The general solution for sin(2x)=5cos^2(x) is x= pi/2+2pin,x=(3pi)/2+2pin,x=1.19028…+pin