What is the general solution for cos(t)-sin(2t)=0 ?

The general solution for cos(t)-sin(2t)=0 is t= pi/2+2pin,t=(3pi)/2+2pin,t= pi/6+2pin,t=(5pi)/6+2pin

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

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

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

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

Rewrite using trig identities

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

Factor

cos (t) - 2 cos (t) sin (t) : - cos (t) (2 sin (t) - 1)

- 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

sin (2 x - \frac{π}{2}) = - 1

3 tan^{2} (2 x) = 1

tan (x) + cot (x) = 3

3 sin (B) - 2 = 5 sin (B) - 1

sin (x) = sin (- x)

What is the general solution for cos(t)-sin(2t)=0 ?
The general solution for cos(t)-sin(2t)=0 is t= pi/2+2pin,t=(3pi)/2+2pin,t= pi/6+2pin,t=(5pi)/6+2pin