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Verify that the \(x\) -values are solutions of the equation. \(3 \tan ^{2} 2 x-1=0\) (a) \(x=\frac{\pi}{12}\) (b) \(x=\frac{5 \pi}{12}\)

Short Answer

Expert verified
Both \(x=\frac{\pi}{12}\) and \(x=\frac{5\pi}{12}\) are solutions of the equation \(3 \tan^2{2x}-1=0\).

Step by step solution

01

Substitute the first given value into the equation

Substitute \(x=\frac{\pi}{12}\) into the given equation, \(3 \tan^2{2x}-1=0\) which leads to \(3 \tan^2{\frac{\pi}{6}}-1=0\)
02

Simplify the equation

Calculate the tangens of \(\frac{\pi}{6}\) (which equals to \(\sqrt{3}/3\)), then square it and multiply by 3, which leads to \(3 * (\frac{\sqrt{3}}{3})^2 - 1= 1 - 1 = 0\).
03

Repeat steps 1 and 2 for the second given value

Substitute \(x=\frac{5\pi}{12}\) into the equation, \(3 \tan^2{2x}-1=0\) which leads to \(3 \tan^2{5\pi/6}-1=0\) And again calculate the tangens, square it and subtract 1. Doing so leads to the equation \(3 * (-\sqrt{3}/3)^2 - 1 = 1 - 1 = 0\)

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