What is the general solution for 0= 20/3 sin(10t)+5cos(10t) ?

The general solution for 0= 20/3 sin(10t)+5cos(10t) is t=-(0.64350…)/(10)+(pin)/(10)

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0= 20/3 sin(10t)+5cos(10t)

0 = \frac{20}{3} sin (10 t) + 5 cos (10 t)

t = - \frac{0.64350 \dots}{10} + \frac{πn}{10}

Solution steps

0 = \frac{20}{3} sin (10 t) + 5 cos (10 t)

Switch sides

\frac{20}{3} sin (10 t) + 5 cos (10 t) = 0

Rewrite using trig identities

\frac{20 tan ( 10 t )}{3} + 5 = 0

Move

5

to the right side

\frac{20 tan ( 10 t )}{3} = - 5

Multiply both sides by

3

20 tan (10 t) = - 15

Divide both sides by

20

tan (10 t) = - \frac{3}{4}

Apply trig inverse properties

10 t = arctan (- \frac{3}{4}) + πn

Solve

10 t = arctan (- \frac{3}{4}) + πn : t = - \frac{arctan ( \frac{3}{4} )}{10} + \frac{πn}{10}

t = - \frac{arctan ( \frac{3}{4} )}{10} + \frac{πn}{10}

Show solutions in decimal form

t = - \frac{0.64350 \dots}{10} + \frac{πn}{10}

y, arctan (y) = \frac{t ^{2}}{2}

cos (- x) = 0

2 tan^{2} (x) + 1 = 0

y, arctan (y) = \frac{x ^{3}}{3} - 2 x + c

cos (A) = \frac{3}{4}

What is the general solution for 0= 20/3 sin(10t)+5cos(10t) ?
The general solution for 0= 20/3 sin(10t)+5cos(10t) is t=-(0.64350…)/(10)+(pin)/(10)