What is the general solution for 3tan^2(x)=2sec^2(x)+1,0<= x<= 180 ?

Q: What is the general solution for 3tan^2(x)=2sec^2(x)+1,0<= x<= 180 ?

The general solution for 3tan^2(x)=2sec^2(x)+1,0<= x<= 180 is x=60,x=120

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Popular Trigonometry >

3tan^2(x)=2sec^2(x)+1,0<= x<= 180

Solution

3 tan^{2} (x) = 2 sec^{2} (x) + 1, 0^{\circ} \leq x \leq 18 0^{\circ}

Solution

x = 6 0^{\circ}, x = 12 0^{\circ}

Radians

x = \frac{π}{3}, x = \frac{2 π}{3}

Solution steps

3 tan^{2} (x) = 2 sec^{2} (x) + 1, 0^{\circ} \leq x \leq 18 0^{\circ}

Subtract

2 sec^{2} (x) + 1

from both sides

3 tan^{2} (x) - 2 sec^{2} (x) - 1 = 0

Rewrite using trig identities

- 1 - 2 sec^{2} (x) + 3 tan^{2} (x)

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

tan^{2} (x) = sec^{2} (x) - 1

= - 1 - 2 sec^{2} (x) + 3 (sec^{2} (x) - 1)

Simplify

- 1 - 2 sec^{2} (x) + 3 (sec^{2} (x) - 1) : sec^{2} (x) - 4

- 1 - 2 sec^{2} (x) + 3 (sec^{2} (x) - 1)

Expand

3 (sec^{2} (x) - 1) : 3 sec^{2} (x) - 3

3 (sec^{2} (x) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = sec^{2} (x), c = 1

= 3 sec^{2} (x) - 3 \cdot 1

Multiply the numbers:

3 \cdot 1 = 3

= 3 sec^{2} (x) - 3

= - 1 - 2 sec^{2} (x) + 3 sec^{2} (x) - 3

Simplify

- 1 - 2 sec^{2} (x) + 3 sec^{2} (x) - 3 : sec^{2} (x) - 4

- 1 - 2 sec^{2} (x) + 3 sec^{2} (x) - 3

Add similar elements:

- 2 sec^{2} (x) + 3 sec^{2} (x) = sec^{2} (x)

= - 1 + sec^{2} (x) - 3

Group like terms

= sec^{2} (x) - 1 - 3

Subtract the numbers:

- 1 - 3 = - 4

= sec^{2} (x) - 4

= sec^{2} (x) - 4

= sec^{2} (x) - 4

- 4 + sec^{2} (x) = 0

Solve by substitution

- 4 + sec^{2} (x) = 0

Let:

sec (x) = u

- 4 + u^{2} = 0

- 4 + u^{2} = 0 : u = 2, u = - 2

- 4 + u^{2} = 0

Move

4

to the right side

- 4 + u^{2} = 0

Add

4

to both sides

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

Simplify

u^{2} = 4

u^{2} = 4

For

x^{2} = f (a)

the solutions are

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

u = 4, u = - 4

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

- 4 = - 2

- 4

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= - 2

u = 2, u = - 2

Substitute back

u = sec (x)

sec (x) = 2, sec (x) = - 2

sec (x) = 2, sec (x) = - 2

sec (x) = 2, 0 \leq x \leq 18 0^{\circ} : x = 6 0^{\circ}

sec (x) = 2, 0 \leq x \leq 18 0^{\circ}

General solutions for

sec (x) = 2

sec (x)

periodicity table with

36 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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solutions for the range

0 \leq x \leq 18 0^{\circ}

x = 6 0^{\circ}

sec (x) = - 2, 0 \leq x \leq 18 0^{\circ} : x = 12 0^{\circ}

sec (x) = - 2, 0 \leq x \leq 18 0^{\circ}

General solutions for

sec (x) = - 2

sec (x)

periodicity table with

36 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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

x = 12 0^{\circ} + 36 0^{\circ} n, x = 24 0^{\circ} + 36 0^{\circ} n

Solutions for the range

0 \leq x \leq 18 0^{\circ}

x = 12 0^{\circ}

Combine all the solutions

x = 6 0^{\circ}, x = 12 0^{\circ}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 3tan^2(x)=2sec^2(x)+1,0<= x<= 180 ?
The general solution for 3tan^2(x)=2sec^2(x)+1,0<= x<= 180 is x=60,x=120