Consider the equation 3tan^2(x)−1=0. the solutions of this equation over the interval {0,pi} are?

If we solve for tan(x) algebraically, we are then able to refer to the unit circle to find the solutions. Solving for tan(x) algebraically will give us tan(x) = (square root of 3)/3. On the unit circle, there are two possibilities with those coordinates: Pi/6 and 7pi/6

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botellok botellok · Answer 2 · 2019-09-01T04:27:52+02:00

Answer:

[tex] x = \frac{\pi}{6}[/tex].

Step-by-step explanation:

We have the following equation:

[tex]3tan^2x - 1 =0[/tex]

[tex]3tan^2x=1[/tex]

[tex]tan^2x=\frac{1}{3}[/tex]

We use the property [tex]tanx = \frac{sinx}{cosx}[/tex] in our equation as follows:

[tex]\frac{sin^2x}{cos^2x}=\frac{1}{3}[/tex]

and then, we use the identity [tex] cos^2x = 1 - sin^2x[/tex]:

[tex]sin^2x=\frac{1}{3}cos^2x[/tex]

[tex]sin^2x=\frac{1}{3}(1- sin^2x)[/tex]

[tex]sin^2x=\frac{1}{3} - \frac{1}{3}sin^2x[/tex]

[tex]sin^2x+\frac{1}{3}sin^2x=\frac{1}{3}[/tex]

[tex]\frac{4}{3}sin^2x = \frac{1}{3}[/tex]

[tex]sin^2x = \frac{1}{4}[/tex]

[tex]sinx = \sqrt{\frac{1}{4}}[/tex]

[tex]sinx =\pm\frac{1}{2}[/tex]

As we are over the interval (0,pi) we use the positive value:

[tex]sinx =\frac{1}{2}[/tex]

[tex]x =sin^{-1}(\frac{1}{2})[/tex]

[tex] x = \frac{\pi}{6}[/tex].