What is the general solution for 5-tan^2(x)=4,0<= x<= 360 ?

Q: What is the general solution for 5-tan^2(x)=4,0<= x<= 360 ?

The general solution for 5-tan^2(x)=4,0<= x<= 360 is x=45,x=225,x=135,x=315

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5-tan^2(x)=4,0<= x<= 360

Solution

5 - tan^{2} (x) = 4, 0^{\circ} \leq x \leq 36 0^{\circ}

Solution

x = 4 5^{\circ}, x = 22 5^{\circ}, x = 13 5^{\circ}, x = 31 5^{\circ}

Radians

x = \frac{π}{4}, x = \frac{5 π}{4}, x = \frac{3 π}{4}, x = \frac{7 π}{4}

Solution steps

5 - tan^{2} (x) = 4, 0^{\circ} \leq x \leq 36 0^{\circ}

Solve by substitution

5 - tan^{2} (x) = 4

Let:

tan (x) = u

5 - u^{2} = 4

5 - u^{2} = 4 : u = 1, u = - 1

5 - u^{2} = 4

Move

5

to the right side

5 - u^{2} = 4

Subtract

5

from both sides

5 - u^{2} - 5 = 4 - 5

Simplify

- u^{2} = - 1

- u^{2} = - 1

Divide both sides by

- 1

- u^{2} = - 1

Divide both sides by

- 1

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

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

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

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = tan (x)

tan (x) = 1, tan (x) = - 1

tan (x) = 1, tan (x) = - 1

tan (x) = 1, 0 \leq x \leq 36 0^{\circ} : x = 4 5^{\circ}, x = 22 5^{\circ}

tan (x) = 1, 0 \leq x \leq 36 0^{\circ}

General solutions for

tan (x) = 1

tan (x)

periodicity table with

18 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 4 5^{\circ} + 18 0^{\circ} n

x = 4 5^{\circ} + 18 0^{\circ} n

Solutions for the range

0 \leq x \leq 36 0^{\circ}

x = 4 5^{\circ}, x = 22 5^{\circ}

tan (x) = - 1, 0 \leq x \leq 36 0^{\circ} : x = 13 5^{\circ}, x = 31 5^{\circ}

tan (x) = - 1, 0 \leq x \leq 36 0^{\circ}

General solutions for

tan (x) = - 1

tan (x)

periodicity table with

18 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = 13 5^{\circ} + 18 0^{\circ} n

x = 13 5^{\circ} + 18 0^{\circ} n

Solutions for the range

0 \leq x \leq 36 0^{\circ}

x = 13 5^{\circ}, x = 31 5^{\circ}

Combine all the solutions

x = 4 5^{\circ}, x = 22 5^{\circ}, x = 13 5^{\circ}, x = 31 5^{\circ}

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Frequently Asked Questions (FAQ)

What is the general solution for 5-tan^2(x)=4,0<= x<= 360 ?
The general solution for 5-tan^2(x)=4,0<= x<= 360 is x=45,x=225,x=135,x=315