2. The function is given by h'(x)=3tan^2x-1 What are the zeros of h on the interval 0leqslant xlt 2pi

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Question

2. The function is given by h'(x)=3tan^2x-1 What are the zeros of h on the interval 0leqslant xlt 2pi

Answer 1

### $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$

Answer 2

## Step 1: Equate the function to zero ### Set $h(x) = 0$: $3 \tan^2 x - 1 = 0$ ## Step 2: Solve for $\tan^2 x$ ### Rewrite equation: $3 \tan^2 x = 1$ ### Solve for $\tan^2 x$: $\tan^2 x = \frac{1}{3}$ ## Step 3: Solve for $\tan x$ ### Take the square root of both sides: $\tan x = \pm \frac{1}{\sqrt{3}}$ ## Step 4: Determine the general solutions ### Use the principal values: $\tan x = \frac{1}{\sqrt{3}}$ and $\tan x = -\frac{1}{\sqrt{3}}$ ### Corresponding angles: $x = \frac{\pi}{6}, \frac{5\pi}{6}$ and $x = \frac{7\pi}{6}, \frac{11\pi}{6}$ ## Step 5: Check interval $0 \leqslant x \lt 2\pi$ ### Verify that all solutions are within the interval

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2. The function is given by h'(x)=3tan^2x-1 What are the zeros of h on the interval 0leqslant xlt 2pi

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