Question

solve the trigonometric equation for all values $0 \leq x < 2\pi$.
$4\cos^2 x - 1 = 0$
answer attempt 1 out of 2
\\(\odot\\) additional solution \\(\odot\\) no solution
$x =$
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Explanation:

Step1: Solve for $\cos^2 x$

Start with the equation \(4\cos^2 x - 1 = 0\). Add 1 to both sides:
\(4\cos^2 x = 1\)
Then divide both sides by 4:
\(\cos^2 x=\frac{1}{4}\)

Step2: Solve for $\cos x$

Take the square root of both sides. Remember that when taking the square root, we consider both positive and negative roots:
\(\cos x=\pm\sqrt{\frac{1}{4}}=\pm\frac{1}{2}\)

Step3: Find $x$ when $\cos x=\frac{1}{2}$

We know that \(\cos x = \frac{1}{2}\) when \(x=\frac{\pi}{3}\) (in the first quadrant) and \(x = 2\pi-\frac{\pi}{3}=\frac{5\pi}{3}\) (in the fourth quadrant) for \(0\leq x<2\pi\).

Step4: Find $x$ when $\cos x=-\frac{1}{2}$

We know that \(\cos x=-\frac{1}{2}\) when \(x=\pi - \frac{\pi}{3}=\frac{2\pi}{3}\) (in the second quadrant) and \(x=\pi+\frac{\pi}{3}=\frac{4\pi}{3}\) (in the third quadrant) for \(0\leq x<2\pi\).

Answer:

The solutions for \(x\) in the interval \(0\leq x < 2\pi\) are \(x=\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}\)