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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

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

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

Solve by substitution

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} : x = 12 0^{\circ}

Combine all the solutions

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

2 arctan (x e^{\frac{0}{2}}) = \frac{π}{2}

2 sin (x) cos (x) - 2 cos (x) = 0

x, 2 cos (x) - 2 = 0

cos (θ) = - \frac{4}{3}

cos (x) = \frac{2.16}{3.42}

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