What is the general solution for (sin(2x))/(cos(x))=2,0<= x<= pi ?

The general solution for (sin(2x))/(cos(x))=2,0<= x<= pi is No Solution for x\in\mathbb{R}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(sin(2x))/(cos(x))=2,0<= x<= pi

\frac{sin ( 2 x )}{cos ( x )} = 2, 0 \leq x \leq π

No Solution for x \in R

Solution steps

\frac{sin ( 2 x )}{cos ( x )} = 2, 0 \leq x \leq π

Subtract

2

from both sides

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

Simplify

\frac{sin ( 2 x )}{cos ( x )} - 2 : \frac{sin ( 2 x ) - 2 cos ( x )}{cos ( x )}

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

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

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

Rewrite using trig identities

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

Factor

- 2 cos (x) + 2 cos (x) sin (x) : 2 cos (x) (sin (x) - 1)

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

Solving each part separately

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

cos (x) = 0, 0 \leq x \leq π : x = \frac{π}{2}

sin (x) - 1 = 0, 0 \leq x \leq π : x = \frac{π}{2}

Combine all the solutions

x = \frac{π}{2}

Since the equation is undefined for:

\frac{π}{2}

No Solution for x \in R

sec (θ) = - 4

sin (x^{2} + 1) = 0

cos (2 x) + sin^{2} (x) = 0

1 + cos (x) = 3 sin (x)

3 sin (x) = 0

What is the general solution for (sin(2x))/(cos(x))=2,0<= x<= pi ?
The general solution for (sin(2x))/(cos(x))=2,0<= x<= pi is No Solution for x\in\mathbb{R}