What is the general solution for sin^2(θ)-tan(θ)cos^2(θ)=0 ?

The general solution for sin^2(θ)-tan(θ)cos^2(θ)=0 is θ=2pin,θ=pi+2pin,θ= pi/4+pin

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Popular >

sin^2(θ)-tan(θ)cos^2(θ)=0

Solution

sin^{2} (θ) - tan (θ) cos^{2} (θ) = 0

Solution

θ = 2 πn, θ = π + 2 πn, θ = \frac{π}{4} + πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n, θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

sin^{2} (θ) - tan (θ) cos^{2} (θ) = 0

Rewrite using trig identities

sin^{2} (θ) - cos^{2} (θ) tan (θ)

Use the basic trigonometric identity:

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

= sin^{2} (θ) - cos^{2} (θ) \frac{sin ( θ )}{cos ( θ )}

cos^{2} (θ) \frac{sin ( θ )}{cos ( θ )} = sin (θ) cos (θ)

cos^{2} (θ) \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{sin ( θ ) cos ^{2} ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= sin (θ) cos (θ)

= sin^{2} (θ) - sin (θ) cos (θ)

sin^{2} (θ) - cos (θ) sin (θ) = 0

Factor

sin^{2} (θ) - cos (θ) sin (θ) : sin (θ) (sin (θ) - cos (θ))

sin^{2} (θ) - cos (θ) sin (θ)

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

sin^{2} (θ) = sin (θ) sin (θ)

= sin (θ) sin (θ) - sin (θ) cos (θ)

Factor out common term

sin (θ)

= sin (θ) (sin (θ) - cos (θ))

sin (θ) (sin (θ) - cos (θ)) = 0

Solving each part separately

sin (θ) = 0 or sin (θ) - cos (θ) = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

General solutions for

sin (θ) = 0

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

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

sin (θ) - cos (θ) = 0

Rewrite using trig identities

sin (θ) - cos (θ) = 0

Divide both sides by

cos (θ), cos (θ) \neq = 0

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

Simplify

\frac{sin ( θ )}{cos ( θ )} - 1 = 0

Use the basic trigonometric identity:

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

tan (θ) - 1 = 0

tan (θ) - 1 = 0

Move

1

to the right side

tan (θ) - 1 = 0

Add

1

to both sides

tan (θ) - 1 + 1 = 0 + 1

Simplify

tan (θ) = 1

tan (θ) = 1

General solutions for

tan (θ) = 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}

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = \frac{π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(θ)-tan(θ)cos^2(θ)=0 ?
The general solution for sin^2(θ)-tan(θ)cos^2(θ)=0 is θ=2pin,θ=pi+2pin,θ= pi/4+pin