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AlekseyPX
2 years ago
12

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

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

Answer:

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

Step-by-step explanation:

rjkz [21]2 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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Answer:

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How many solutions are there in this equation 8 (x-3)=8x-24
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Answer:

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1 year ago
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The equation of a parabola is y=x^2-10x+27. Write the equation in vertex form
azamat

Answer:

\large \boxed{y = (x - 5)^{2} + 2 }

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\begin{array}{rcll}y & = & x^{2} - 10x + 27 & \\y - 27 & = & x^{2} - 10x & \text{Subtracted 27 from each side}\\y - 27&= & x^{2} - 10x + 25 - 25 & \text{Added and subtracted (b/2)}^{2}\\y - 27&= & (x - 5)^{2} - 25 & \text{Wrote the first three terms as the square of a binomial}\\y& = & \mathbf{(x - 5)^{2} + 2} & \text{Added 27 to each side}\\\end{array}\\\text{The vertex form of the parabola is $\large \boxed{\mathbf{y = (x - 5)^{2} + 2 }}$}The figure below shows that your parabola has its vertex at (5,2).

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Answer:

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2 years ago
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