<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: Third Option

Step-by-step explanation:
We have the following exponential equation

We must solve the equation for the variable x
To clear the variable x apply the
function on both sides of the equation

Simplifying we get the following:

To simplify the expression
we apply the base change property

This means that:

Then:



Step-by-step explanation:
follow the above attachment, hope this helps you.
Answer:
minus 7a plus 11
Step-by-step explanation:
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For Question Number one
(2+10)²+4=148
Question Number two:
(6+5)×(8-6)=22