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Popular >

(tan(θ)-2)(16sin^2(θ)-1)=0

Solution

(tan (θ) - 2) (16 sin^{2} (θ) - 1) = 0

Solution

θ = 1.10714 \dots + πn, θ = 0.25268 \dots + 2 πn, θ = π - 0.25268 \dots + 2 πn, θ = - 0.25268 \dots + 2 πn, θ = π + 0.25268 \dots + 2 πn

Degrees

θ = 63.43494 \dots^{\circ} + 18 0^{\circ} n, θ = 14.47751 \dots^{\circ} + 36 0^{\circ} n, θ = 165.52248 \dots^{\circ} + 36 0^{\circ} n, θ = - 14.47751 \dots^{\circ} + 36 0^{\circ} n, θ = 194.47751 \dots^{\circ} + 36 0^{\circ} n

Solution steps

(tan (θ) - 2) (16 sin^{2} (θ) - 1) = 0

Solving each part separately

tan (θ) - 2 = 0 or 16 sin^{2} (θ) - 1 = 0

tan (θ) - 2 = 0 : θ = arctan (2) + πn

tan (θ) - 2 = 0

Move

2

to the right side

tan (θ) - 2 = 0

Add

2

to both sides

tan (θ) - 2 + 2 = 0 + 2

Simplify

tan (θ) = 2

tan (θ) = 2

Apply trig inverse properties

tan (θ) = 2

General solutions for

tan (θ) = 2

tan (x) = a \Rightarrow x = arctan (a) + πn

θ = arctan (2) + πn

θ = arctan (2) + πn

16 sin^{2} (θ) - 1 = 0 : θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn, θ = arcsin (- \frac{1}{4}) + 2 πn, θ = π + arcsin (\frac{1}{4}) + 2 πn

16 sin^{2} (θ) - 1 = 0

Solve by substitution

16 sin^{2} (θ) - 1 = 0

Let:

sin (θ) = u

16 u^{2} - 1 = 0

16 u^{2} - 1 = 0 : u = \frac{1}{4}, u = - \frac{1}{4}

16 u^{2} - 1 = 0

Move

1

to the right side

16 u^{2} - 1 = 0

Add

1

to both sides

16 u^{2} - 1 + 1 = 0 + 1

Simplify

16 u^{2} = 1

16 u^{2} = 1

Divide both sides by

16

16 u^{2} = 1

Divide both sides by

16

\frac{16 u ^{2}}{16} = \frac{1}{16}

Simplify

u^{2} = \frac{1}{16}

u^{2} = \frac{1}{16}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{1}{16}, u = - \frac{1}{16}

\frac{1}{16} = \frac{1}{4}

\frac{1}{16}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{16}

16 = 4

16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

= \frac{1}{4}

Apply rule

1 = 1

= \frac{1}{4}

- \frac{1}{16} = - \frac{1}{4}

- \frac{1}{16}

Simplify

\frac{1}{16} : \frac{1}{4}

\frac{1}{16}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{16}

16 = 4

16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

= \frac{1}{4}

= - \frac{1}{4}

Apply rule

1 = 1

= - \frac{1}{4}

u = \frac{1}{4}, u = - \frac{1}{4}

Substitute back

u = sin (θ)

sin (θ) = \frac{1}{4}, sin (θ) = - \frac{1}{4}

sin (θ) = \frac{1}{4}, sin (θ) = - \frac{1}{4}

sin (θ) = \frac{1}{4} : θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

sin (θ) = \frac{1}{4}

Apply trig inverse properties

sin (θ) = \frac{1}{4}

General solutions for

sin (θ) = \frac{1}{4}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn

sin (θ) = - \frac{1}{4} : θ = arcsin (- \frac{1}{4}) + 2 πn, θ = π + arcsin (\frac{1}{4}) + 2 πn

sin (θ) = - \frac{1}{4}

Apply trig inverse properties

sin (θ) = - \frac{1}{4}

General solutions for

sin (θ) = - \frac{1}{4}

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

θ = arcsin (- \frac{1}{4}) + 2 πn, θ = π + arcsin (\frac{1}{4}) + 2 πn

θ = arcsin (- \frac{1}{4}) + 2 πn, θ = π + arcsin (\frac{1}{4}) + 2 πn

Combine all the solutions

θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn, θ = arcsin (- \frac{1}{4}) + 2 πn, θ = π + arcsin (\frac{1}{4}) + 2 πn

Combine all the solutions

θ = arctan (2) + πn, θ = arcsin (\frac{1}{4}) + 2 πn, θ = π - arcsin (\frac{1}{4}) + 2 πn, θ = arcsin (- \frac{1}{4}) + 2 πn, θ = π + arcsin (\frac{1}{4}) + 2 πn

Show solutions in decimal form

θ = 1.10714 \dots + πn, θ = 0.25268 \dots + 2 πn, θ = π - 0.25268 \dots + 2 πn, θ = - 0.25268 \dots + 2 πn, θ = π + 0.25268 \dots + 2 πn

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