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

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

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

- 2 sin (t) + cos (2 t) \cdot 2 = 0

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

Solution steps

- 2 sin (t) + cos (2 t) \cdot 2 = 0

Rewrite using trig identities

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

Solve by substitution

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

sin (t) = - 1 : t = \frac{3 π}{2} + 2 πn

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

Combine all the solutions

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

tan (a) = \frac{4}{3}

5 sin (θ) - 1 = 0

tan (x) csc (x) - tan (x) = 0

cos (6 0^{\circ} - x) = sin (70 - 3 x)

sin (- x) = sin (x)

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