What is the general solution for 25tan^2(θ)-4=0 ?

The general solution for 25tan^2(θ)-4=0 is θ=0.38050…+pin,θ=-0.38050…+pin

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Popular >

25tan^2(θ)-4=0

Solution

25 tan^{2} (θ) - 4 = 0

Solution

θ = 0.38050 \dots + πn, θ = - 0.38050 \dots + πn

Degrees

θ = 21.80140 \dots^{\circ} + 18 0^{\circ} n, θ = - 21.80140 \dots^{\circ} + 18 0^{\circ} n

Solution steps

25 tan^{2} (θ) - 4 = 0

Solve by substitution

25 tan^{2} (θ) - 4 = 0

Let:

tan (θ) = u

25 u^{2} - 4 = 0

25 u^{2} - 4 = 0 : u = \frac{2}{5}, u = - \frac{2}{5}

25 u^{2} - 4 = 0

Move

4

to the right side

25 u^{2} - 4 = 0

Add

4

to both sides

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

Simplify

25 u^{2} = 4

25 u^{2} = 4

Divide both sides by

25

25 u^{2} = 4

Divide both sides by

25

\frac{25 u ^{2}}{25} = \frac{4}{25}

Simplify

u^{2} = \frac{4}{25}

u^{2} = \frac{4}{25}

For

x^{2} = f (a)

the solutions are

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

u = \frac{4}{25}, u = - \frac{4}{25}

\frac{4}{25} = \frac{2}{5}

\frac{4}{25}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{4}{25}

25 = 5

25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

= \frac{4}{5}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{2}{5}

- \frac{4}{25} = - \frac{2}{5}

- \frac{4}{25}

Simplify

\frac{4}{25} : \frac{2}{5}

\frac{4}{25}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{4}{25}

25 = 5

25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

= \frac{4}{5}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{2}{5}

= - \frac{2}{5}

u = \frac{2}{5}, u = - \frac{2}{5}

Substitute back

u = tan (θ)

tan (θ) = \frac{2}{5}, tan (θ) = - \frac{2}{5}

tan (θ) = \frac{2}{5}, tan (θ) = - \frac{2}{5}

tan (θ) = \frac{2}{5} : θ = arctan (\frac{2}{5}) + πn

tan (θ) = \frac{2}{5}

Apply trig inverse properties

tan (θ) = \frac{2}{5}

General solutions for

tan (θ) = \frac{2}{5}

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

θ = arctan (\frac{2}{5}) + πn

θ = arctan (\frac{2}{5}) + πn

tan (θ) = - \frac{2}{5} : θ = arctan (- \frac{2}{5}) + πn

tan (θ) = - \frac{2}{5}

Apply trig inverse properties

tan (θ) = - \frac{2}{5}

General solutions for

tan (θ) = - \frac{2}{5}

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

θ = arctan (- \frac{2}{5}) + πn

θ = arctan (- \frac{2}{5}) + πn

Combine all the solutions

θ = arctan (\frac{2}{5}) + πn, θ = arctan (- \frac{2}{5}) + πn

Show solutions in decimal form

θ = 0.38050 \dots + πn, θ = - 0.38050 \dots + πn

Popular Examples

tan(x)= 8/15

tan (x) = \frac{8}{15}

sin(θ)=-12/13

sin (θ) = - \frac{12}{13}

sin(x)cos(x)=-1/4

sin (x) cos (x) = - \frac{1}{4}

sec^2(θ)-9=0

sec^{2} (θ) - 9 = 0

2cos^2(θ)=2-sin(θ)

2 cos^{2} (θ) = 2 - sin (θ)

Frequently Asked Questions (FAQ)

What is the general solution for 25tan^2(θ)-4=0 ?
The general solution for 25tan^2(θ)-4=0 is θ=0.38050…+pin,θ=-0.38050…+pin