What is the general solution for 5tan^2(x)-7=8 ?

The general solution for 5tan^2(x)-7=8 is x= pi/3+pin,x=(2pi)/3+pin

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5tan^2(x)-7=8

5 tan^{2} (x) - 7 = 8

x = \frac{π}{3} + πn, x = \frac{2 π}{3} + πn

Degrees

x = 6 0^{\circ} + 18 0^{\circ} n, x = 12 0^{\circ} + 18 0^{\circ} n

Solution steps

5 tan^{2} (x) - 7 = 8

Solve by substitution

5 tan^{2} (x) - 7 = 8

Let:

tan (x) = u

5 u^{2} - 7 = 8

5 u^{2} - 7 = 8 : u = 3, u = - 3

5 u^{2} - 7 = 8

Move

7

to the right side

5 u^{2} - 7 = 8

Add

7

to both sides

5 u^{2} - 7 + 7 = 8 + 7

Simplify

5 u^{2} = 15

5 u^{2} = 15

Divide both sides by

5

5 u^{2} = 15

Divide both sides by

5

\frac{5 u ^{2}}{5} = \frac{15}{5}

Simplify

u^{2} = 3

u^{2} = 3

For

x^{2} = f (a)

the solutions are

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

u = 3, u = - 3

Substitute back

u = tan (x)

tan (x) = 3, tan (x) = - 3

tan (x) = 3, tan (x) = - 3

tan (x) = 3 : x = \frac{π}{3} + πn

tan (x) = 3

General solutions for

tan (x) = 3

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{3} + πn

x = \frac{π}{3} + πn

tan (x) = - 3 : x = \frac{2 π}{3} + πn

tan (x) = - 3

General solutions for

tan (x) = - 3

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{2 π}{3} + πn

x = \frac{2 π}{3} + πn

Combine all the solutions

x = \frac{π}{3} + πn, x = \frac{2 π}{3} + πn

3 sin (2 y) - 4 sin (2 y) cos (2 y) = 0

sin (x) + sin (4 x) = 0

cos (0.568 x - 2.595) = 0

8 tan (- x + \frac{π}{8}) = 8

tan (\frac{A}{2}) = 0.62

What is the general solution for 5tan^2(x)-7=8 ?
The general solution for 5tan^2(x)-7=8 is x= pi/3+pin,x=(2pi)/3+pin