What is the general solution for sin(2t)cos(t)=0,0<= t<= pi ?

The general solution for sin(2t)cos(t)=0,0<= t<= pi is t=0,t= pi/2 ,t=pi

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sin(2t)cos(t)=0,0<= t<= pi

sin (2 t) cos (t) = 0, 0 \leq t \leq π

t = 0, t = \frac{π}{2}, t = π

Degrees

t = 0^{\circ}, t = 9 0^{\circ}, t = 18 0^{\circ}

Solution steps

sin (2 t) cos (t) = 0, 0 \leq t \leq π

Solving each part separately

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

sin (2 t) = 0, 0 \leq t \leq π : t = 0, t = \frac{π}{2}, t = π

sin (2 t) = 0, 0 \leq t \leq π

General solutions for

sin (2 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}

2 t = 0 + 2 πn, 2 t = π + 2 πn

2 t = 0 + 2 πn, 2 t = π + 2 πn

Solve

2 t = 0 + 2 πn : t = πn

2 t = 0 + 2 πn

0 + 2 πn = 2 πn

2 t = 2 πn

Divide both sides by

2

2 t = 2 πn

Divide both sides by

2

\frac{2 t}{2} = \frac{2 πn}{2}

Simplify

t = πn

t = πn

Solve

2 t = π + 2 πn : t = \frac{π}{2} + πn

2 t = π + 2 πn

Divide both sides by

2

2 t = π + 2 πn

Divide both sides by

2

\frac{2 t}{2} = \frac{π}{2} + \frac{2 πn}{2}

Simplify

t = \frac{π}{2} + πn

t = \frac{π}{2} + πn

t = πn, t = \frac{π}{2} + πn

Solutions for the range

0 \leq t \leq π

t = 0, t = \frac{π}{2}, t = π

cos (t) = 0, 0 \leq t \leq π : t = \frac{π}{2}

cos (t) = 0, 0 \leq t \leq π

General solutions for

cos (t) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

Solutions for the range

0 \leq t \leq π

t = \frac{π}{2}

Combine all the solutions

t = 0, t = \frac{π}{2}, t = π

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

θ, y = 4 sin (θ)

- 3 (sin^{2} (θ) - cos^{2} (θ)) = 0

cos^{2} (t) - sin^{2} (t) = - 1

- a^{2} c_{1} cos (a x) - a^{2} c_{2} sin (a x) = 0

What is the general solution for sin(2t)cos(t)=0,0<= t<= pi ?
The general solution for sin(2t)cos(t)=0,0<= t<= pi is t=0,t= pi/2 ,t=pi