Problem 50 - Algebra and Trigonometry

Question

Involve trigonometric equations quadratic in form. Solve each equation on the interval \([0,2 \pi)\) $$ \sec ^{2} x-2=0 $$

Step-by-Step Solution

Answer

The solutions for the original equation \(\sec^2 x - 2 = 0\) within the interval [0, 2π) are \(x = \frac{\pi}{4}, \frac{7\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}\).

Step 1: Rewrite secant as reciprocal of cosine

Rewrite the secant (\(\sec\)) function in the equation as the reciprocal of cosine (\(\cos\)). This gives: \(\frac{1}{\cos^2x} - 2 = 0\).

Step 2: Solve for \(\cos^2x\)

Isolate \(\cos^2x\) on one side of the equation: \(\cos^2x = \frac{1}{2}\).

Step 3: Take square root on both sides

Since \(\cos^2x\) is equal to \(\frac{1}{2}\), taking the square root on both sides gives us \(\cos x = ±\sqrt{\frac{1}{2}} = ±\frac{1}{\sqrt{2}} = ±\frac{\sqrt{2}}{2}\).

Step 4: Find all solutions on the interval [0, 2π)

The solutions in the interval [0, 2π) for \(\cos x = \frac{\sqrt{2}}{2}\) are \(x = \frac{\pi}{4}, \frac{7\pi}{4}\) and for \(\cos x = -\frac{\sqrt{2}}{2}\) are \(x = \frac{3\pi}{4}, \frac{5\pi}{4}\).

Step 5: Compilation of solutions

So, all the solutions for the original equation \(\sec^2 x - 2 = 0\) within the interval [0, 2π) are \(x = \frac{\pi}{4}, \frac{7\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}\).