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rewona [7]
3 years ago
7

Solve the following equation on the interval [0, 2π). tan^2x sin x = tan^2x

Mathematics
1 answer:
Alenkasestr [34]3 years ago
8 0
<h3>Given</h3>

tan(x)²·sin(x) = tan(x)²

<h3>Find</h3>

x on the interval [0, 2π)

<h3>Solution</h3>

Subtract the right side and factor. Then make use of the zero-product rule.

... tan(x)²·sin(x) -tan(x)² = 0

... tan(x)²·(sin(x) -1) = 0

This is an indeterminate form at x = π/2 and undefined at x = 3π/2. We can resolve the indeterminate form by using an identity for tan(x)²:

... tan(x)² = sin(x)²/cos(x)² = sin(x)²/(1 -sin(x)²)

Then our equation becomes

... sin(x)²·(sin(x) -1)/((1 -sin(x))(1 +sin(x))) = 0

... -sin(x)²/(1 +sin(x)) = 0

Now, we know the only solutions are found where sin(x) = 0, at ...

... x ∈ {0, π}

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On the other hand, if x = 0 then y = b.

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C. y = 5x+1, y = -3x+4
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</span>
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Finally the answer is B. So, the inequalities are:  

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