What is the general solution for csc(θ)cos(θ)sin(θ)=1 ?

The general solution for csc(θ)cos(θ)sin(θ)=1 is No Solution for θ\in\mathbb{R}

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csc(θ)cos(θ)sin(θ)=1

csc (θ) cos (θ) sin (θ) = 1

No Solution for θ \in R

Solution steps

csc (θ) cos (θ) sin (θ) = 1

Subtract

1

from both sides

csc (θ) cos (θ) sin (θ) - 1 = 0

Rewrite using trig identities

- 1 + cos (θ) csc (θ) sin (θ)

Use the basic trigonometric identity:

sin (x) = \frac{1}{csc ( x )}

= - 1 + cos (θ) csc (θ) \frac{1}{csc ( θ )}

cos (θ) csc (θ) \frac{1}{csc ( θ )} = cos (θ)

cos (θ) csc (θ) \frac{1}{csc ( θ )}

Multiply fractions:

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

= \frac{1 \cdot cos ( θ ) csc ( θ )}{csc ( θ )}

Cancel the common factor:

csc (θ)

= 1 \cdot cos (θ)

Multiply:

1 \cdot cos (θ) = cos (θ)

= cos (θ)

= - 1 + cos (θ)

- 1 + cos (θ) = 0

Move

1

to the right side

- 1 + cos (θ) = 0

Add

1

to both sides

- 1 + cos (θ) + 1 = 0 + 1

Simplify

cos (θ) = 1

cos (θ) = 1

General solutions for

cos (θ) = 1

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 = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

Since the equation is undefined for:

2 πn

No Solution for θ \in R

3 cos (x) + sin (2 x) = 0

tan (α) = 0.875

sin (x) = \frac{18}{31}

4 cos^{3} (x) - 3 cos (x) = - 0.7071

2 tan (x) = tan (2 x)

What is the general solution for csc(θ)cos(θ)sin(θ)=1 ?
The general solution for csc(θ)cos(θ)sin(θ)=1 is No Solution for θ\in\mathbb{R}