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

The general solution for sin(t)+cos(2t)=0 is t=(7pi)/6+2pin,t=(11pi)/6+2pin,t= pi/2+2pin

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

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

t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn, t = \frac{π}{2} + 2 πn

Solution steps

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

Rewrite using trig identities

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

Solve by substitution

sin (t) = - \frac{1}{2}, sin (t) = 1

sin (t) = - \frac{1}{2} : t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn

sin (t) = 1 : t = \frac{π}{2} + 2 πn

Combine all the solutions

t = \frac{7 π}{6} + 2 πn, t = \frac{11 π}{6} + 2 πn, t = \frac{π}{2} + 2 πn

cos (θ) = \frac{2}{3}, \frac{sin ( θ )}{2}, 27 0^{\circ} < θ < 36 0^{\circ}

cos (9 x) = 0

2 cos^{3} (x) - 3 sin^{2} (x) cos (x) = 0

5 cot (θ) = - 6

cos (3 x - 9 x) = 0

What is the general solution for sin(t)+cos(2t)=0 ?
The general solution for sin(t)+cos(2t)=0 is t=(7pi)/6+2pin,t=(11pi)/6+2pin,t= pi/2+2pin