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

The general solution for csc(θ)sin(θ)=cot(θ) is θ= pi/4+pin

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csc(θ)sin(θ)=cot(θ)

csc (θ) sin (θ) = cot (θ)

θ = \frac{π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

csc (θ) sin (θ) = cot (θ)

Rewrite using trig identities

csc (θ) sin (θ)

csc (θ) sin (θ) = 1

csc (θ) sin (θ)

Express with sin, cos

csc (θ) sin (θ)

Use the basic trigonometric identity:

csc (θ) = \frac{1}{sin ( θ )}

= \frac{1}{sin ( θ )} sin (θ)

\frac{1}{sin ( θ )} sin (θ) = 1

\frac{1}{sin ( θ )} sin (θ)

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{1 sin ( θ )}{sin ( θ )}

Cancel the common factor:

sin (θ)

= 1

= 1

= 1

1 = cot (θ)

Switch sides

cot (θ) = 1

General solutions for

cot (θ) = 1

cot (x)

periodicity table with

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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

7 sin (θ) + cos (θ) = 5

tan (θ + π) + 2 sin (θ + π) = 0

sin (θ) = \frac{2}{3}, \frac{π}{2} < θ < π, cos (2 θ) = x

tan (3 x) = 2

tan (θ) = 3 - π

What is the general solution for csc(θ)sin(θ)=cot(θ) ?
The general solution for csc(θ)sin(θ)=cot(θ) is θ= pi/4+pin