What is the general solution for solvefor ϕ,(xcos(ϕ))^2=x^2sin^2(ϕ)(1) ?

The general solution for solvefor ϕ,(xcos(ϕ))^2=x^2sin^2(ϕ)(1) is ϕ=(3pi)/4+pin,ϕ= pi/4+pin

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solvefor ϕ,(xcos(ϕ))^2=x^2sin^2(ϕ)(1)

Solution

solve for

ϕ, (x cos (ϕ))^{2} = x^{2} sin^{2} (ϕ) (1)

Solution

ϕ = \frac{3 π}{4} + πn, ϕ = \frac{π}{4} + πn

Degrees

ϕ = 13 5^{\circ} + 18 0^{\circ} n, ϕ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

(x cos (ϕ))^{2} = x^{2} sin^{2} (ϕ) (1)

Subtract

x^{2} sin^{2} (ϕ) 1

from both sides

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

Factor

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

x^{2} cos^{2} (ϕ) - x^{2} sin^{2} (ϕ)

Rewrite

x^{2} cos^{2} (ϕ)

(x cos (ϕ))^{2}

x^{2} cos^{2} (ϕ)

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

x^{2} cos^{2} (ϕ) = (x cos (ϕ))^{2}

= (x cos (ϕ))^{2}

Rewrite

x^{2} sin^{2} (ϕ)

(x sin (ϕ))^{2}

x^{2} sin^{2} (ϕ)

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

x^{2} sin^{2} (ϕ) = (x sin (ϕ))^{2}

= (x sin (ϕ))^{2}

= (x cos (ϕ))^{2} - (x sin (ϕ))^{2}

Apply Difference of Two Squares Formula:

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

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

= (x cos (ϕ) + x sin (ϕ)) (x cos (ϕ) - x sin (ϕ))

Factor

x cos (ϕ) + x sin (ϕ) : x (cos (ϕ) + sin (ϕ))

x cos (ϕ) + x sin (ϕ)

Factor out common term

x

= x (cos (ϕ) + sin (ϕ))

= x (cos (ϕ) + sin (ϕ)) (x cos (ϕ) - x sin (ϕ))

Factor

x cos (ϕ) - x sin (ϕ) : x (cos (ϕ) - sin (ϕ))

x cos (ϕ) - x sin (ϕ)

Factor out common term

x

= x (cos (ϕ) - sin (ϕ))

= x (cos (ϕ) + sin (ϕ)) x (cos (ϕ) - sin (ϕ))

Refine

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

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

Solving each part separately

cos (ϕ) + sin (ϕ) = 0 or cos (ϕ) - sin (ϕ) = 0

cos (ϕ) + sin (ϕ) = 0 : ϕ = \frac{3 π}{4} + πn

cos (ϕ) + sin (ϕ) = 0

Rewrite using trig identities

cos (ϕ) + sin (ϕ) = 0

Divide both sides by

cos (ϕ), cos (ϕ) \neq = 0

\frac{cos ( ϕ ) + sin ( ϕ )}{cos ( ϕ )} = \frac{0}{cos ( ϕ )}

Simplify

1 + \frac{sin ( ϕ )}{cos ( ϕ )} = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

1 + tan (ϕ) = 0

1 + tan (ϕ) = 0

Move

1

to the right side

1 + tan (ϕ) = 0

Subtract

1

from both sides

1 + tan (ϕ) - 1 = 0 - 1

Simplify

tan (ϕ) = - 1

tan (ϕ) = - 1

General solutions for

tan (ϕ) = - 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

ϕ = \frac{3 π}{4} + πn

ϕ = \frac{3 π}{4} + πn

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

cos (ϕ) - sin (ϕ) = 0

Rewrite using trig identities

cos (ϕ) - sin (ϕ) = 0

Divide both sides by

cos (ϕ), cos (ϕ) \neq = 0

\frac{cos ( ϕ ) - sin ( ϕ )}{cos ( ϕ )} = \frac{0}{cos ( ϕ )}

Simplify

1 - \frac{sin ( ϕ )}{cos ( ϕ )} = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

1 - tan (ϕ) = 0

1 - tan (ϕ) = 0

Move

1

to the right side

1 - tan (ϕ) = 0

Subtract

1

from both sides

1 - tan (ϕ) - 1 = 0 - 1

Simplify

- tan (ϕ) = - 1

- tan (ϕ) = - 1

Divide both sides by

- 1

- tan (ϕ) = - 1

Divide both sides by

- 1

\frac{- tan ( ϕ )}{- 1} = \frac{- 1}{- 1}

Simplify

tan (ϕ) = 1

tan (ϕ) = 1

General solutions for

tan (ϕ) = 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

ϕ = \frac{π}{4} + πn

ϕ = \frac{π}{4} + πn

Combine all the solutions

ϕ = \frac{3 π}{4} + πn, ϕ = \frac{π}{4} + πn

Popular Examples

cos(180-x)=sin(-300)

cos (18 0^{\circ} - x) = sin (- 30 0^{\circ})

sin^2(x)=((6m-5))/8

sin^{2} (x) = \frac{( 6 m - 5 )}{8}

cos(θ)=0.417

cos (θ) = 0.417

tan(x)=2.7475

tan (x) = 2.7475

cot(a)=1

cot (a) = 1

Frequently Asked Questions (FAQ)

What is the general solution for solvefor ϕ,(xcos(ϕ))^2=x^2sin^2(ϕ)(1) ?
The general solution for solvefor ϕ,(xcos(ϕ))^2=x^2sin^2(ϕ)(1) is ϕ=(3pi)/4+pin,ϕ= pi/4+pin