What is the general solution for sin(x)=sec(x) ?

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

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sin(x)=sec(x)

sin (x) = sec (x)

No Solution for x \in R

Solution steps

sin (x) = sec (x)

Subtract

sec (x)

from both sides

sin (x) - sec (x) = 0

Express with sin, cos

- sec (x) + sin (x)

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

m, f = e^{n x} cos (m x)

4 sin^{4} (x) - 8 sin^{2} (x) + 3 = 0

sin (x - \frac{π}{3}) = 0.4

2 sin^{2} (θ) - 2 = cos^{2} (θ)

sin (θ) + cos (θ) = \frac{3}{2}

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