The answer it 1. is A.
The answer to 2. is B.
I assume you're asked to solve
4 cos²(<em>x</em>) - 7 cos(<em>x</em>) + 3 = 0
Factor the left side:
(4 cos(<em>x</em>) - 3) (cos(<em>x</em>) - 1) = 0
Then either
4 cos(<em>x</em>) - 3 = 0 <u>or</u> cos(<em>x</em>) - 1 = 0
cos(<em>x</em>) = 3/4 <u>or</u> cos(<em>x</em>) = 1
From the first case, we get
<em>x</em> = cos⁻¹(3/4) + 2<em>nπ</em> <u>or</u> <em>x</em> = -cos⁻¹(3/4) + 2<em>nπ</em>
and from the second,
<em>x</em> = <em>nπ</em>
where <em>n</em> is any integer.
Answer: A
Step-by-step explanation: true true true
Answer:
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Step-by-step explanation:
1) The Fundamental Theorem of Calculus in its first part, shows us a reciprocal relationship between Derivatives and Integration
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2) In this case, we'll need to find the derivative applying the chain rule. As it follows:

3) To test it, just integrate:

The answer to your question is y = sin(x + 90) as we have a sine graph phase shifted 90 units left