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AlekseyPX
3 years ago
12

Solve 2cos x+2cos 2x=0 on the interval [0,2pi)

Mathematics
2 answers:
Anarel [89]3 years ago
8 0

Answer:

The answers are B and C, 3π/4 and 5π/4

Step-by-step explanation:

rjkz [21]3 years ago
6 0
2cosx+2cos2x=0 \\ -2sin^2x+2cos^2x+2cosx=0 \\ -2(1-cos^2x)+2cos^2x+2cosx=0 \\ -2+2cos^2x+2cos^2x+2cosx=0 \\ 4cos^2x+2cosx-2=0 \\ t=cos x \\ 4t^2+2t-2=0 \\ \Delta=4+32=36 \\ t_1= \frac{-2-6}{8}=-1 \\ t_2= \frac{-2+6}{8}= \frac{1}{2}  \\ cos x=-1 \lor cos x= \frac{1}{2}    \\ x=-\pi \lor x= \frac{5\pi}{3}
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The difference of the quotient m and 9 and the quotient of n and 30
Zanzabum
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First we start with m/9-n/30.

We multiply the denominators to get 270.

Then we multiply the numerators by the original denominators to get (30m-9n).

To check, we can use m=2, and n=4

2/9-4/30=4/45

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3 years ago
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If f(x) = -7x + 5x2 + 10, what does f(-2) equal?
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Answer:

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Step-by-step explanation:

Hello there! Lets get started solving this function!

For a function, whenever we see f(x) = ... , the function is simply asking the value of the equation at x! So, if replace f(x) with f(-2), we simply substitute x with -2 in the equation! Lets get started!

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