What is the general solution for cos(2x)=sin(x-pi/4) ?

The general solution for cos(2x)=sin(x-pi/4) is x= pi/4+pin,x=2pin+(11pi)/(12),x=2pin+(19pi)/(12)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(2x)=sin(x-pi/4)

cos (2 x) = sin (x - \frac{π}{4})

x = \frac{π}{4} + πn, x = 2 πn + \frac{11 π}{12}, x = 2 πn + \frac{19 π}{12}

Solution steps

cos (2 x) = sin (x - \frac{π}{4})

Rewrite using trig identities

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

Subtract

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

from both sides

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

Simplify

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

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

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

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

Rewrite using trig identities

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

Factor

cos (x) - sin (x) + (cos^{2} (x) - sin^{2} (x)) 2 : (cos (x) - sin (x)) (2 (cos (x) + sin (x)) + 1)

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

Solving each part separately

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

cos (x) - sin (x) = 0 : x = \frac{π}{4} + πn

2 (cos (x) + sin (x)) + 1 = 0 : x = 2 πn + \frac{11 π}{12}, x = 2 πn + \frac{19 π}{12}

Combine all the solutions

x = \frac{π}{4} + πn, x = 2 πn + \frac{11 π}{12}, x = 2 πn + \frac{19 π}{12}

sin (2 x + 60) + sin (x + 30) = 0, 0 \leq x \leq 2 π

tan^{2} (x) = 3 - csc^{2} (x)

2 sin^{2} (u) = 1 + sin (u)

2 tan^{2} (θ) - 3 tan (θ) + 1 = 0

cos (x) = \frac{9}{17}

What is the general solution for cos(2x)=sin(x-pi/4) ?
The general solution for cos(2x)=sin(x-pi/4) is x= pi/4+pin,x=2pin+(11pi)/(12),x=2pin+(19pi)/(12)