What is the general solution for 4sin(2t)=4cos(t) ?

The general solution for 4sin(2t)=4cos(t) is t= pi/2+2pin,t=(3pi)/2+2pin,t= pi/6+2pin,t=(5pi)/6+2pin

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4sin(2t)=4cos(t)

4 sin (2 t) = 4 cos (t)

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn, t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

Solution steps

4 sin (2 t) = 4 cos (t)

Subtract

4 cos (t)

from both sides

4 sin (2 t) - 4 cos (t) = 0

Rewrite using trig identities

- 4 cos (t) + 8 cos (t) sin (t) = 0

Factor

- 4 cos (t) + 8 cos (t) sin (t) : 4 cos (t) (2 sin (t) - 1)

4 cos (t) (2 sin (t) - 1) = 0

Solving each part separately

cos (t) = 0 or 2 sin (t) - 1 = 0

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

2 sin (t) - 1 = 0 : t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

Combine all the solutions

t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn, t = \frac{π}{6} + 2 πn, t = \frac{5 π}{6} + 2 πn

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

1 + 2 sin^{2} (t) = - 5 cos (t)

2 sin (2 x) - 3 - \frac{6}{sin ( 2 x ) - 1} = 0

3 sin^{2} (x) + 11 sin (x) + 10 = 0

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

What is the general solution for 4sin(2t)=4cos(t) ?
The general solution for 4sin(2t)=4cos(t) is t= pi/2+2pin,t=(3pi)/2+2pin,t= pi/6+2pin,t=(5pi)/6+2pin