What is the general solution for 3tan^2(x)=1,-pi<= x<= pi ?

The general solution for 3tan^2(x)=1,-pi<= x<= pi is x= pi/6 ,x=-(5pi)/6 ,x=-pi/6 ,x=(5pi)/6

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3tan^2(x)=1,-pi<= x<= pi

3 tan^{2} (x) = 1, - π \leq x \leq π

x = \frac{π}{6}, x = - \frac{5 π}{6}, x = - \frac{π}{6}, x = \frac{5 π}{6}

Degrees

x = 3 0^{\circ}, x = - 15 0^{\circ}, x = - 3 0^{\circ}, x = 15 0^{\circ}

Solution steps

3 tan^{2} (x) = 1, - π \leq x \leq π

Solve by substitution

3 tan^{2} (x) = 1

Let:

tan (x) = u

3 u^{2} = 1

3 u^{2} = 1 : u = \frac{1}{3}, u = - \frac{1}{3}

3 u^{2} = 1

Divide both sides by

3

3 u^{2} = 1

Divide both sides by

3

\frac{3 u ^{2}}{3} = \frac{1}{3}

Simplify

u^{2} = \frac{1}{3}

u^{2} = \frac{1}{3}

For

x^{2} = f (a)

the solutions are

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

u = \frac{1}{3}, u = - \frac{1}{3}

Substitute back

u = tan (x)

tan (x) = \frac{1}{3}, tan (x) = - \frac{1}{3}

tan (x) = \frac{1}{3}, tan (x) = - \frac{1}{3}

tan (x) = \frac{1}{3}, - π \leq x \leq π : x = \frac{π}{6}, x = - \frac{5 π}{6}

tan (x) = \frac{1}{3}, - π \leq x \leq π

Apply trig inverse properties

tan (x) = \frac{1}{3}

General solutions for

tan (x) = \frac{1}{3}

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

x = arctan (\frac{1}{3}) + πn

x = arctan (\frac{1}{3}) + πn

Solutions for the range

- π \leq x \leq π

x = \frac{π}{6}, x = - \frac{5 π}{6}

tan (x) = - \frac{1}{3}, - π \leq x \leq π : x = - \frac{π}{6}, x = \frac{5 π}{6}

tan (x) = - \frac{1}{3}, - π \leq x \leq π

Apply trig inverse properties

tan (x) = - \frac{1}{3}

General solutions for

tan (x) = - \frac{1}{3}

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

x = arctan (- \frac{1}{3}) + πn

x = arctan (- \frac{1}{3}) + πn

Solutions for the range

- π \leq x \leq π

x = - \frac{π}{6}, x = \frac{5 π}{6}

Combine all the solutions

x = \frac{π}{6}, x = - \frac{5 π}{6}, x = - \frac{π}{6}, x = \frac{5 π}{6}

cos (2 t) - cos (t) = 0.5

0 = - 6 csc (x) cot (x)

0.08 = 0.12 cos^{2} (3.922 t)

\frac{15}{sin ( 3 2 ^{\circ} )} = \frac{12}{sin ( b )}

\frac{sin ( x )}{25} = \frac{sin ( 12 5 ^{\circ} )}{165.6}

What is the general solution for 3tan^2(x)=1,-pi<= x<= pi ?
The general solution for 3tan^2(x)=1,-pi<= x<= pi is x= pi/6 ,x=-(5pi)/6 ,x=-pi/6 ,x=(5pi)/6