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

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

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sin(θ)cos(θ)=cos(θ)

sin (θ) cos (θ) = cos (θ)

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

Degrees

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

Solution steps

sin (θ) cos (θ) = cos (θ)

Subtract

cos (θ)

from both sides

sin (θ) cos (θ) - cos (θ) = 0

Factor

sin (θ) cos (θ) - cos (θ) : cos (θ) (sin (θ) - 1)

sin (θ) cos (θ) - cos (θ)

Factor out common term

cos (θ)

= cos (θ) (sin (θ) - 1)

cos (θ) (sin (θ) - 1) = 0

Solving each part separately

cos (θ) = 0 or sin (θ) - 1 = 0

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

cos (θ) = 0

General solutions for

cos (θ) = 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}

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

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

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

sin (θ) - 1 = 0

Move

1

to the right side

sin (θ) - 1 = 0

Add

1

to both sides

sin (θ) - 1 + 1 = 0 + 1

Simplify

sin (θ) = 1

sin (θ) = 1

General solutions for

sin (θ) = 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}

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

Combine all the solutions

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

tan^{2} (2 θ) = 3

tan^{2} (2 θ) = 1

(50.691) cos (50.69 1^{\circ} - c) + sin (50.69 1^{\circ} - c) = 0

sec (4 θ) = 2

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

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