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

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

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cos(x)=csc(x)

cos (x) = csc (x)

No Solution for x \in R

Solution steps

cos (x) = csc (x)

Subtract

csc (x)

from both sides

cos (x) - csc (x) = 0

Express with sin, cos

cos (x) - csc (x)

Use the basic trigonometric identity:

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

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

Simplify

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

cos (x) - \frac{1}{sin ( x )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

- 1 + cos (x) sin (x) = 0

Rewrite using trig identities

- 1 + cos (x) sin (x)

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 x )}{2}

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

Move

1

to the right side

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

Add

1

to both sides

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

Simplify

\frac{sin ( 2 x )}{2} = 1

\frac{sin ( 2 x )}{2} = 1

Multiply both sides by

2

\frac{sin ( 2 x )}{2} = 1

Multiply both sides by

2

\frac{2 sin ( 2 x )}{2} = 1 \cdot 2

Simplify

sin (2 x) = 2

sin (2 x) = 2

- 1 \leq sin (x) \leq 1

No Solution for x \in R

1 + cos (2 t) = cos^{2} (t)

2 sin (2 x + \frac{3 π}{2}) + 1 = 0

2 cos (2 x) - sin (x) - 1 = 0

2 cos^{2} (θ) - 2 sin^{2} (θ) = 2

cot (θ) + 3 csc (θ) = 5

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