Question

question 15 (5 points)
cos 2x + \frac{1}{2} = 0
\frac{\pi}{3} + 2\pi n; \frac{2\pi}{3} + 2\pi n
\frac{\pi}{6} + \pi n; \frac{\pi}{3} + \pi n
\frac{\pi}{6} + 2\pi n; \frac{\pi}{3} + 2\pi n
\frac{\pi}{3} + \pi n; \frac{2\pi}{3} + \pi n
question 16 (5 points)
simplify \frac{\sin x}{\sec x + 1}
cot x - \cot x\cos x
cot x + \cot x\cos x
cot x - \cot x\sin x
\sin x\cos x + \sin x

Explanation:

Question 15

Step1: Solve for \(\cos2x\)

Given \(\cos2x+\frac{1}{2} = 0\), we can rewrite it as \(\cos2x=-\frac{1}{2}\).

Step2: Find the general solution for \(2x\)

We know that if \(\cos\theta =-\frac{1}{2}\), then \(\theta=\frac{2\pi}{3}+2k\pi\) or \(\theta = \frac{4\pi}{3}+2k\pi\), \(k\in\mathbb{Z}\). Substituting \(\theta = 2x\), we get \(2x=\frac{2\pi}{3}+2k\pi\) or \(2x=\frac{4\pi}{3}+2k\pi\).

Step3: Solve for \(x\)

Dividing each equation by \(2\), we have \(x=\frac{\pi}{3}+k\pi\) or \(x=\frac{2\pi}{3}+k\pi\), \(k\in\mathbb{Z}\)

Step1: Rewrite \(\sec x\)

Since \(\sec x=\frac{1}{\cos x}\), the expression \(\frac{\sin x}{\sec x + 1}\) becomes \(\frac{\sin x}{\frac{1}{\cos x}+1}=\frac{\sin x\cos x}{1 + \cos x}\)

Step2: Use the identity \(\cot x=\frac{\cos x}{\sin x}\)

Multiply numerator and denominator by \(\sin x\) (not necessary, another approach: \(\frac{\sin x\cos x}{1+\cos x}=\frac{\cos x(1 - \cos^{2}x)}{(1 + \cos x)\sin x}\) (using \(\sin^{2}x=1-\cos^{2}x\)). Simplify \(\frac{\cos x(1 - \cos x)(1+\cos x)}{(1 + \cos x)\sin x}=\frac{\cos x(1 - \cos x)}{\sin x}=\cot x-\cot x\cos x\)

Answer:

\(\frac{\pi}{3}+\pi n;\frac{2\pi}{3}+\pi n\) (the fourth option)

Question 16