Solve the trigonometric equation for all values $$ 0 \leq x

Question

Solve the trigonometric equation for all values $$ 0 \leq x<2 \pi $$ $$ \sec x-2=0 $$

UpStudy AI · Accepted Answer

We start with the equation
$$
\sec x - 2 = 0.
$$

1. Add 2 to both sides:
$$
\sec x = 2.
$$

2. Recall that the secant function is the reciprocal of the cosine function, so:
$$
\sec x = \frac{1}{\cos x}.
$$
This gives us:
$$
\frac{1}{\cos x} = 2.
$$

3. Solve for $$ \cos x $$ by taking the reciprocal of both sides:
$$
\cos x = \frac{1}{2}.
$$

4. We now need to find all values of $$ x $$ in the interval $$0 \leq x < 2\pi$$ for which:
$$
\cos x = \frac{1}{2}.
$$
The cosine function is positive in the first and fourth quadrants. The reference angle $$ \alpha $$ satisfying:
$$
\cos \alpha = \frac{1}{2}
$$
is:
$$
\alpha = \frac{\pi}{3}.
$$

5. Therefore, the two solutions in the interval $$0 \leq x < 2\pi$$ are:
$$
x = \frac{\pi}{3} \quad \text{and} \quad x = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}.
$$

Thus, the solutions are:
$$
x = \frac{\pi}{3} \quad \text{and} \quad x = \frac{5\pi}{3}.
$$
The solutions are $$ x = \frac{\pi}{3} $$ and $$ x = \frac{5\pi}{3} $$.

Solve the trigonometric equation for all values \( 0 \leq x<2 \pi \) \[ \sec x-2=0 \]

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