What is the general solution for -1.9sin(t)=0.4cos(2t) ?

The general solution for -1.9sin(t)=0.4cos(2t) is t=-0.19583…+2pin,t=pi+0.19583…+2pin

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-1.9sin(t)=0.4cos(2t)

- 1.9 sin (t) = 0.4 cos (2 t)

t = - 0.19583 \dots + 2 πn, t = π + 0.19583 \dots + 2 πn

Solution steps

- 1.9 sin (t) = 0.4 cos (2 t)

Subtract

0.4 cos (2 t)

from both sides

- 1.9 sin (t) - 0.4 cos (2 t) = 0

Rewrite using trig identities

- (1 - 2 sin^{2} (t)) \cdot 0.4 - 1.9 sin (t) = 0

Solve by substitution

sin (t) = \frac{19 + 489}{16}, sin (t) = \frac{19 - 489}{16}

sin (t) = \frac{19 + 489}{16} :

No Solution

sin (t) = \frac{19 - 489}{16} : t = arcsin (\frac{19 - 489}{16}) + 2 πn, t = π + arcsin (- \frac{19 - 489}{16}) + 2 πn

Combine all the solutions

t = arcsin (\frac{19 - 489}{16}) + 2 πn, t = π + arcsin (- \frac{19 - 489}{16}) + 2 πn

Show solutions in decimal form

t = - 0.19583 \dots + 2 πn, t = π + 0.19583 \dots + 2 πn

sin (\frac{x}{2}) = \frac{2}{2}

sin (2 t) = 1

sin^{4} (x) = sin^{2} (x)

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

0.32 = 0.59406 cos (x)

What is the general solution for -1.9sin(t)=0.4cos(2t) ?
The general solution for -1.9sin(t)=0.4cos(2t) is t=-0.19583…+2pin,t=pi+0.19583…+2pin