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

The general solution for 4tan^2(θ)-25=0 is θ=1.19028…+pin,θ=-1.19028…+pin

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Popular >

4tan^2(θ)-25=0

Solution

4 tan^{2} (θ) - 25 = 0

Solution

θ = 1.19028 \dots + πn, θ = - 1.19028 \dots + πn

Degrees

θ = 68.19859 \dots^{\circ} + 18 0^{\circ} n, θ = - 68.19859 \dots^{\circ} + 18 0^{\circ} n

Solution steps

4 tan^{2} (θ) - 25 = 0

Solve by substitution

4 tan^{2} (θ) - 25 = 0

Let:

tan (θ) = u

4 u^{2} - 25 = 0

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

4 u^{2} - 25 = 0

Move

25

to the right side

4 u^{2} - 25 = 0

Add

25

to both sides

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

Simplify

4 u^{2} = 25

4 u^{2} = 25

Divide both sides by

4

4 u^{2} = 25

Divide both sides by

4

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

Simplify

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

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

For

x^{2} = f (a)

the solutions are

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

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

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

\frac{25}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{25}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{25}{2}

25 = 5

25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

= \frac{5}{2}

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

- \frac{25}{4}

Simplify

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

\frac{25}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{25}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{25}{2}

25 = 5

25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

= \frac{5}{2}

= - \frac{5}{2}

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

Substitute back

u = tan (θ)

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

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

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

tan (θ) = \frac{5}{2}

Apply trig inverse properties

tan (θ) = \frac{5}{2}

General solutions for

tan (θ) = \frac{5}{2}

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

θ = arctan (\frac{5}{2}) + πn

θ = arctan (\frac{5}{2}) + πn

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

tan (θ) = - \frac{5}{2}

Apply trig inverse properties

tan (θ) = - \frac{5}{2}

General solutions for

tan (θ) = - \frac{5}{2}

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

θ = arctan (- \frac{5}{2}) + πn

θ = arctan (- \frac{5}{2}) + πn

Combine all the solutions

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

Show solutions in decimal form

θ = 1.19028 \dots + πn, θ = - 1.19028 \dots + πn

Popular Examples

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sin^{2} (x) - cos^{2} (x) = - 1

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4sin(θ)cos^2(θ)=3sin(θ)

4 sin (θ) cos^{2} (θ) = 3 sin (θ)

sin(x)= 7/25

sin (x) = \frac{7}{25}

cos(x)+9sin^2(x)=1

cos (x) + 9 sin^{2} (x) = 1

Frequently Asked Questions (FAQ)

What is the general solution for 4tan^2(θ)-25=0 ?
The general solution for 4tan^2(θ)-25=0 is θ=1.19028…+pin,θ=-1.19028…+pin