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)

solve for

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

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

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 (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 : ϕ = \frac{π}{4} + πn

Combine all the solutions

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

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

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

cos (θ) = 0.417

tan (x) = 2.7475

cot (a) = 1

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