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3 years ago

7

Solve cos x + rad 2 = -cos x for over the interval [0, 2pi ]

1 answer:

Kamila [148]3 years ago

6 0

$cos x + \sqrt{2} = -cosx \\ \\ 2 cos x + \sqrt{2} = 0 \\ \\ cos x = -\frac{\sqrt{2}}{2}$

The cosine function is negative in 2nd and 3rd quadrants, So you know that you will have 2 solutions. One between pi/2 and pi, the other between pi and 3pi/2.

Refer to a unit circle to find that
$cos (\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2} \\ \\ cos (\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2}$

Final Answer:
$x = \frac{3\pi}{4}, \frac{5\pi}{4}$

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hoa [83]

Answer:

a) This integral can be evaluated using the basic integration rules. $\int 11x^{4}dx = \frac{11}{5} x^{5}+C$

b) This integral can be evaluated using the basic integration rules. $\int 8x^{1}x^{4}dx=\frac{4}{3}x^{6}+C$

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Step-by-step explanation:

a) $\int 11x^{4}dx$

In order to solve this problem, we can directly make use of the power rule of integration, which looks like this:

$\int kx^{n}=k\frac{x^{n+1}}{n+1}+C$

so in this case we would get:

$\int 11x^{4}dx=11 \frac{x^{4+1}}{4+1}+C$

$\int 11x^{4}dx=11 \frac{x^{5}}{5}+C$

b) $\int 8x^{1}x^{4}dx$

In order to solve this problem we just need to use some algebra to simplify it. By using power rules, we get that:

$\int 8x^{1}x^{4}dx=\int 8x^{1+4}dx=\int 8x^{5}dx$

So we can now use the power rule of integration:

$\int 8x^{5}dx=\frac{8}{5+1}x^{5+1}+C$

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$\int 8x^{5}dx=\frac{4}{3}x^{6}+C$

c) The same applies to this problem:

$\int 3x^{31}x^{4}dx=\int 3x^{31+4}dx=\int 3x^{35}dx$

and now we can use the power rule of integration:

$\int 3x^{35}dx=\frac{3x^{35+1}}{35+1}+C$

$\int 3x^{35}dx=\frac{3x^{36}}{36}+C$

$\int 3x^{35}dx=\frac{x^{36}}{12}+C$

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