What is the general solution for -10sin^2(t)=0 ?

The general solution for -10sin^2(t)=0 is t=2pin,t=pi+2pin

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-10sin^2(t)=0

- 10 sin^{2} (t) = 0

t = 2 πn, t = π + 2 πn

Degrees

t = 0^{\circ} + 36 0^{\circ} n, t = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

- 10 sin^{2} (t) = 0

Divide both sides by

- 10

- 10 sin^{2} (t) = 0

Divide both sides by

- 10

- 10 sin^{2} (t) = 0

Divide both sides by

- 10

\frac{- 10 sin ^{2} ( t )}{- 10} = \frac{0}{- 10}

Simplify

sin^{2} (t) = 0

sin^{2} (t) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

sin (t) = 0

General solutions for

sin (t) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

t = 0 + 2 πn, t = π + 2 πn

t = 0 + 2 πn, t = π + 2 πn

Solve

t = 0 + 2 πn : t = 2 πn

t = 0 + 2 πn

0 + 2 πn = 2 πn

t = 2 πn

t = 2 πn, t = π + 2 πn

x, e^{y} cos (2 x) + y = 1

cos (θ) = 0.4, sec (θ)

- \frac{π}{2} sin (\frac{π x}{2}) = 0

cot (θ) = \frac{1 + cos ^{2} ( θ )}{2 sin ( θ ) cos ( θ )}

cot (x) = 3 + 1

What is the general solution for -10sin^2(t)=0 ?
The general solution for -10sin^2(t)=0 is t=2pin,t=pi+2pin