Determine the zeros and state the multiplicity of any repeated zeros for f(x)=3x(x+2)^2(2x-1)^3
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<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}
Answer:
∠J = 60°
Step-by-step explanation:
The Law of Cosines tells you ...
j² = k² +l² -2kl·cos(J)
Solving for J gives ...
J = arccos((k² +l² -j²)/(2kl))
J = arccos((14² +80² -74²)/(2·14·80)) = arccos(1120/2240) = arccos(1/2)
J = 60°
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<em>Additional comment</em>
It is pretty rare to find a set of integer side lengths that result in one of the angles of the triangle being a rational number of degrees.
