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

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

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cos(θ)=csc(θ)

cos (θ) = csc (θ)

No Solution for θ \in R

Solution steps

cos (θ) = csc (θ)

Subtract

csc (θ)

from both sides

cos (θ) - csc (θ) = 0

Express with sin, cos

cos (θ) - csc (θ)

Use the basic trigonometric identity:

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

= cos (θ) - \frac{1}{sin ( θ )}

Simplify

cos (θ) - \frac{1}{sin ( θ )} : \frac{cos ( θ ) sin ( θ ) - 1}{sin ( θ )}

cos (θ) - \frac{1}{sin ( θ )}

Convert element to fraction:

cos (θ) = \frac{cos ( θ ) sin ( θ )}{sin ( θ )}

= \frac{cos ( θ ) sin ( θ )}{sin ( θ )} - \frac{1}{sin ( θ )}

Since the denominators are equal, combine the fractions:

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

= \frac{cos ( θ ) sin ( θ ) - 1}{sin ( θ )}

= \frac{cos ( θ ) sin ( θ ) - 1}{sin ( θ )}

\frac{- 1 + cos ( θ ) sin ( θ )}{sin ( θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- 1 + cos (θ) sin (θ) = 0

Rewrite using trig identities

- 1 + cos (θ) sin (θ)

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

sin (x) cos (x) = \frac{sin ( 2 x )}{2}

= - 1 + \frac{sin ( 2 θ )}{2}

- 1 + \frac{sin ( 2 θ )}{2} = 0

Move

1

to the right side

- 1 + \frac{sin ( 2 θ )}{2} = 0

Add

1

to both sides

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

Simplify

\frac{sin ( 2 θ )}{2} = 1

\frac{sin ( 2 θ )}{2} = 1

Multiply both sides by

2

\frac{sin ( 2 θ )}{2} = 1

Multiply both sides by

2

\frac{2 sin ( 2 θ )}{2} = 1 \cdot 2

Simplify

sin (2 θ) = 2

sin (2 θ) = 2

- 1 \leq sin (x) \leq 1

No Solution for θ \in R

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

cos (2 x) + cos (x) = 2

sec^{2} (θ) + 2 tan (θ) = 0

4 sin^{2} (\frac{x}{2}) + 8 cos (\frac{x}{2}) - 7 = 0

sin (x) + cos (x) = 1, - 36 0^{\circ} \leq x \leq 36 0^{\circ}

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