What is the general solution for sin(t)cos(t)=cos(t) ?

The general solution for sin(t)cos(t)=cos(t) is t= pi/2+2pin,t=(3pi)/2+2pin

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sin(t)cos(t)=cos(t)

sin (t) cos (t) = cos (t)

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

Degrees

t = 9 0^{\circ} + 36 0^{\circ} n, t = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (t) cos (t) = cos (t)

Subtract

cos (t)

from both sides

sin (t) cos (t) - cos (t) = 0

Factor

sin (t) cos (t) - cos (t) : cos (t) (sin (t) - 1)

sin (t) cos (t) - cos (t)

Factor out common term

cos (t)

= cos (t) (sin (t) - 1)

cos (t) (sin (t) - 1) = 0

Solving each part separately

cos (t) = 0 or sin (t) - 1 = 0

cos (t) = 0 : t = \frac{π}{2} + 2 πn, t = \frac{3 π}{2} + 2 πn

cos (t) = 0

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

sin (t) - 1 = 0 : t = \frac{π}{2} + 2 πn

sin (t) - 1 = 0

Move

1

to the right side

sin (t) - 1 = 0

Add

1

to both sides

sin (t) - 1 + 1 = 0 + 1

Simplify

sin (t) = 1

sin (t) = 1

General solutions for

sin (t) = 1

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 = \frac{π}{2} + 2 πn

t = \frac{π}{2} + 2 πn

Combine all the solutions

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

- 7 cos (7 x) = 0

1 - 4 sin^{2} (x) = 0

cos (2 θ) - 6 sin^{2} (θ) = - 5

sin (A) = \frac{15}{17}

9 sin (x) = 0

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