What is the general solution for sin(φ)-cos(φ)=sin(φ)tan(φ) ?

The general solution for sin(φ)-cos(φ)=sin(φ)tan(φ) is No Solution for φ\in\mathbb{R}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(φ)-cos(φ)=sin(φ)tan(φ)

sin (φ) - cos (φ) = sin (φ) tan (φ)

No Solution for φ \in R

Solution steps

sin (φ) - cos (φ) = sin (φ) tan (φ)

Subtract

sin (φ) tan (φ)

from both sides

sin (φ) - cos (φ) - sin (φ) tan (φ) = 0

Express with sin, cos

\frac{- cos ^{2} ( φ ) - sin ^{2} ( φ ) + cos ( φ ) sin ( φ )}{cos ( φ )} = 0

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

- cos^{2} (φ) - sin^{2} (φ) + cos (φ) sin (φ) = 0

Rewrite using trig identities

- 1 + \frac{sin ( 2 φ )}{2} = 0

Move

1

to the right side

\frac{sin ( 2 φ )}{2} = 1

Multiply both sides by

2

sin (2 φ) = 2

- 1 \leq sin (x) \leq 1

No Solution for φ \in R

cos (x) - cos (\frac{x}{2}) = 0

sin (x) = 0, - \frac{π}{4} < x < \frac{9 π}{4}

2 tan (x) cos (x) = tan (x)

sin (x) = \frac{5}{4}

sin (2 θ) = 3 cos (θ)

What is the general solution for sin(φ)-cos(φ)=sin(φ)tan(φ) ?
The general solution for sin(φ)-cos(φ)=sin(φ)tan(φ) is No Solution for φ\in\mathbb{R}