What is the general solution for 9tan^2(θ)-4=0 ?

The general solution for 9tan^2(θ)-4=0 is θ=0.58800…+pin,θ=-0.58800…+pin

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Popular >

9tan^2(θ)-4=0

Solution

9 tan^{2} (θ) - 4 = 0

Solution

θ = 0.58800 \dots + πn, θ = - 0.58800 \dots + πn

Degrees

θ = 33.69006 \dots^{\circ} + 18 0^{\circ} n, θ = - 33.69006 \dots^{\circ} + 18 0^{\circ} n

Solution steps

9 tan^{2} (θ) - 4 = 0

Solve by substitution

9 tan^{2} (θ) - 4 = 0

Let:

tan (θ) = u

9 u^{2} - 4 = 0

9 u^{2} - 4 = 0 : u = \frac{2}{3}, u = - \frac{2}{3}

9 u^{2} - 4 = 0

Move

4

to the right side

9 u^{2} - 4 = 0

Add

4

to both sides

9 u^{2} - 4 + 4 = 0 + 4

Simplify

9 u^{2} = 4

9 u^{2} = 4

Divide both sides by

9

9 u^{2} = 4

Divide both sides by

9

\frac{9 u ^{2}}{9} = \frac{4}{9}

Simplify

u^{2} = \frac{4}{9}

u^{2} = \frac{4}{9}

For

x^{2} = f (a)

the solutions are

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

u = \frac{4}{9}, u = - \frac{4}{9}

\frac{4}{9} = \frac{2}{3}

\frac{4}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{4}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{4}{3}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{2}{3}

- \frac{4}{9} = - \frac{2}{3}

- \frac{4}{9}

Simplify

\frac{4}{9} : \frac{2}{3}

\frac{4}{9}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{4}{9}

9 = 3

9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

= \frac{4}{3}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{2}{3}

= - \frac{2}{3}

u = \frac{2}{3}, u = - \frac{2}{3}

Substitute back

u = tan (θ)

tan (θ) = \frac{2}{3}, tan (θ) = - \frac{2}{3}

tan (θ) = \frac{2}{3}, tan (θ) = - \frac{2}{3}

tan (θ) = \frac{2}{3} : θ = arctan (\frac{2}{3}) + πn

tan (θ) = \frac{2}{3}

Apply trig inverse properties

tan (θ) = \frac{2}{3}

General solutions for

tan (θ) = \frac{2}{3}

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

θ = arctan (\frac{2}{3}) + πn

θ = arctan (\frac{2}{3}) + πn

tan (θ) = - \frac{2}{3} : θ = arctan (- \frac{2}{3}) + πn

tan (θ) = - \frac{2}{3}

Apply trig inverse properties

tan (θ) = - \frac{2}{3}

General solutions for

tan (θ) = - \frac{2}{3}

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

θ = arctan (- \frac{2}{3}) + πn

θ = arctan (- \frac{2}{3}) + πn

Combine all the solutions

θ = arctan (\frac{2}{3}) + πn, θ = arctan (- \frac{2}{3}) + πn

Show solutions in decimal form

θ = 0.58800 \dots + πn, θ = - 0.58800 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 9tan^2(θ)-4=0 ?
The general solution for 9tan^2(θ)-4=0 is θ=0.58800…+pin,θ=-0.58800…+pin