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3 years ago

Find all the solutions in the interval [0,2π) sin²x = 2-2cosX

Mathematics

1 answer:

Finger [1]3 years ago

5 0

Step-by-step explanation:

sin²x = 2 - 2cosx

1 - cos²x = 2 - 2cosx (because sin²x + cos²x = 1)

cos²x - 2cosx + 1 = 0

(cosx - 1)² = 0

cosx - 1 = 0

cosx = 1

Hence x = 0.

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Calculate the flux of the vector field F⃗ (x,y,z)=(exy+9z+4)i⃗ +(exy+4z+9)j⃗ +(9z+exy)k⃗ through the square of side length 3 wit

ikadub [295]

The square (call it $S$ ) has one vertex at the origin (0, 0, 0) and one edge on the y-axis, which tells us another vertex is (0, 3, 0). The normal vector to the plane is $\vec n=\vec\imath-\vec k$ , which is enough information to figure out the equation of the plane containing $S$ :

$(x\,\vec\imath+y\,\vec\jmath+z\,\vec k)\cdot(\vec\imath-\vec k)=0\implies x-z=0\implies z=x$

We can parameterize this surface by

$\vec s(x,y)=x\,\vec\imath+y\,\vec\jmath+x\,\vec k$

for $0\le x\le\frac3{\sqrt2}$ and $0\le y\le3$ . Then the flux of $\vec F$ , assumed to be

$\vec F(x,y,z)=(e^{xy}+9z+4)\,\vec\imath+(e^{xy}+4z+9)\,\vec\jmath+(9ze^{xy})\,\vec k$ ,

$\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\iint_S\vec F(\vec s(x,y))\cdot\vec n\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_0^3\int_0^{3/\sqrt2}\left((4+e^{xy}+9x)\,\vec\imath+(9+e^{xy}+4x)\,\vec\jmath+(e^{xy}+9x)\,\vec k\right)\cdot(\vec\imath-\vec k)\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_0^3\int_0^{3/\sqrt2}4\,\mathrm dx\,\mathrm dy=\boxed{18\sqrt2}$

3 0

3 years ago

RJ read 40 pages of gis book in 50 minutes. How many pages can I read in 80 minutes?

lisov135 [29]

Answer:

64 pages

Step-by-step explanation:

40 pg = 50 min

8 pg = 10 min

64 pg = 80 min

5 0

3 years ago

How do you do this problem? I need to know how you found the answer.

Alexxandr [17]

to get the equation of a line, we simply need two points, say for the Red one ... notice in the graph the lines passes through (0,2) and (-1,6), so let's use those

$\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{-1-0}\implies \cfrac{4}{-1}\implies -4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=-4(x-0) \\\\\\ y-2=-4x\implies \blacktriangleright y=-4x+2 \blacktriangleleft$

now, for the Blue one, say let's use hmmm it passes through (0,2) and (1.6)

$\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{1-0}\implies \cfrac{4}{1}\implies 4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=4(x-0) \\\\\\ y-2=4x\implies \blacktriangleright y=4x+2 \blacktriangleleft$

3 0

3 years ago

Write a problem that can be solved by dividing 9/10 by 1/4

exis [7]

Answer:

Step-by-step explanation:

8 0

3 years ago

Randy assembles bookcases as a part-time job. He is responsible for assembling two different sizes of bookcases.

Aleonysh [2.5K]

Answer: The Answer to this is

The line representing the number of screws required to assemble x of Bookcase B is steeper than the line representing the number of screws required to assemble x of Bookcase A because there are more screws used in each Bookcase B than in each Bookcase A.

Step-by-step explanation:

<h2>This is a thing I wish other people should do so here's the rest of the three answers. the answer for −3.1□−2.4 is <u> </u><u><em><</em></u>. The answer for −3/5□−2/3 is <u>> </u>, and the answer for this number 4 is <u>The line representing Bicyclist A is steeper than the line representing Bicyclist B, therefore Bicyclist B is moving at a slower rate.</u></h2><h2>I hope this helps.</h2><h2><u> </u></h2><h2><u> </u></h2><h2><u> </u></h2><h2><u> </u></h2><h2><u /></h2>

8 0

2 years ago