What is the general solution for solvefor x,y=cos(4x) ?

The general solution for solvefor x,y=cos(4x) is x=(arccos(y))/4+(pin)/2 ,x=-(arccos(y))/4+(pin)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor x,y=cos(4x)

solve for

x, y = cos (4 x)

x = \frac{arccos ( y )}{4} + \frac{πn}{2}, x = - \frac{arccos ( y )}{4} + \frac{πn}{2}

Solution steps

y = cos (4 x)

Switch sides

cos (4 x) = y

Apply trig inverse properties

cos (4 x) = y

General solutions for

cos (4 x) = y

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = - arccos (a) + 2 πn

4 x = arccos (y) + 2 πn, 4 x = - arccos (y) + 2 πn

4 x = arccos (y) + 2 πn, 4 x = - arccos (y) + 2 πn

Solve

4 x = arccos (y) + 2 πn : x = \frac{arccos ( y )}{4} + \frac{πn}{2}

4 x = arccos (y) + 2 πn

Divide both sides by

4

4 x = arccos (y) + 2 πn

Divide both sides by

4

\frac{4 x}{4} = \frac{arccos ( y )}{4} + \frac{2 πn}{4}

Simplify

x = \frac{arccos ( y )}{4} + \frac{πn}{2}

x = \frac{arccos ( y )}{4} + \frac{πn}{2}

Solve

4 x = - arccos (y) + 2 πn : x = - \frac{arccos ( y )}{4} + \frac{πn}{2}

4 x = - arccos (y) + 2 πn

Divide both sides by

4

4 x = - arccos (y) + 2 πn

Divide both sides by

4

\frac{4 x}{4} = - \frac{arccos ( y )}{4} + \frac{2 πn}{4}

Simplify

x = - \frac{arccos ( y )}{4} + \frac{πn}{2}

x = - \frac{arccos ( y )}{4} + \frac{πn}{2}

x = \frac{arccos ( y )}{4} + \frac{πn}{2}, x = - \frac{arccos ( y )}{4} + \frac{πn}{2}

tan^{4} (x) - 2 tan^{2} (x) - 1 = 0

- 0.002 = sin (2 π \cdot (3) - 500 π (0) + x)

3 tan (2 x - \frac{π}{2}) = 2 sin (\frac{π}{3})

2 tan (x) - 2 = 0, 0 \leq x \leq 2 π

tan (3 5^{\circ}) tan (5 6^{\circ} - x) = 1

What is the general solution for solvefor x,y=cos(4x) ?
The general solution for solvefor x,y=cos(4x) is x=(arccos(y))/4+(pin)/2 ,x=-(arccos(y))/4+(pin)/2