What is the general solution for solvefor x,-y^2-cos(x)=0 ?

The general solution for solvefor x,-y^2-cos(x)=0 is x=arccos(-y^2)+2pin,x=-arccos(-y^2)+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor x,-y^2-cos(x)=0

solve for

x, - y^{2} - cos (x) = 0

x = arccos (- y^{2}) + 2 πn, x = - arccos (- y^{2}) + 2 πn

Solution steps

- y^{2} - cos (x) = 0

Move

y^{2}

to the right side

- y^{2} - cos (x) = 0

Add

y^{2}

to both sides

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

Simplify

- cos (x) = y^{2}

- cos (x) = y^{2}

Divide both sides by

- 1

- cos (x) = y^{2}

Divide both sides by

- 1

\frac{- cos ( x )}{- 1} = \frac{y ^{2}}{- 1}

Simplify

cos (x) = - y^{2}

cos (x) = - y^{2}

Apply trig inverse properties

cos (x) = - y^{2}

General solutions for

cos (x) = - y^{2}

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

x = arccos (- y^{2}) + 2 πn, x = - arccos (- y^{2}) + 2 πn

x = arccos (- y^{2}) + 2 πn, x = - arccos (- y^{2}) + 2 πn

cos (θ) = \frac{6}{16}

∣ cos (x) ∣ = - 1

\frac{sin ( 2 5 ^{\circ} )}{5.41} = \frac{sin ( x )}{9}

cos (θ) = \frac{4}{17}

- csc^{2} (x) + 2 csc (x) cot (x) = 0

What is the general solution for solvefor x,-y^2-cos(x)=0 ?
The general solution for solvefor x,-y^2-cos(x)=0 is x=arccos(-y^2)+2pin,x=-arccos(-y^2)+2pin