What is the general solution for 0=-2sin(2x)+cos(2x) ?

The general solution for 0=-2sin(2x)+cos(2x) is x=(0.46364…)/2+(pin)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

x = \frac{0.46364 \dots}{2} + \frac{πn}{2}

Solution steps

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

Switch sides

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

Rewrite using trig identities

- 2 tan (2 x) + 1 = 0

Move

1

to the right side

- 2 tan (2 x) = - 1

Divide both sides by

- 2

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

Apply trig inverse properties

2 x = arctan (\frac{1}{2}) + πn

Solve

2 x = arctan (\frac{1}{2}) + πn : x = \frac{arctan ( \frac{1}{2} )}{2} + \frac{πn}{2}

x = \frac{arctan ( \frac{1}{2} )}{2} + \frac{πn}{2}

Show solutions in decimal form

x = \frac{0.46364 \dots}{2} + \frac{πn}{2}

y, arcsin (y) = arcsin (x)

4 csc (x) + 2 sin (x) = 9

x, h = 1 + sin (π x)

1 - cos (θ) = sin (\frac{θ}{2})

cos (\frac{θ}{3} + 2 4^{\circ}) = sin (θ)

What is the general solution for 0=-2sin(2x)+cos(2x) ?
The general solution for 0=-2sin(2x)+cos(2x) is x=(0.46364…)/2+(pin)/2