Question

Solve each of the following. Please give both the general solution and the specific solutions in the given interval. You may copy this page and put it into a word document then you can type or write directly on this sheet and attach to assignment 0.07.

Answer 1

http://learn.flvs.net/webdav/educator_calculus_v8/module00/0_07a.htm

Answer 2

@John_ES

Answer 3

For the first question,
\[\sin2\theta=-\frac{\sqrt{3}}{2}\]\[\Rightarrow 2\theta_1=\frac{4\pi}{3}\Rightarrow \theta_1=2\pi/3+2\pi k\]\[\Rightarrow 2\theta_2=\frac{5\pi}{3}\Rightarrow \theta_2=5\pi/6+2\pi k\]

Answer 4

For the second question,
\[2\sin^2x=2+\cos x\Rightarrow 2(1-\cos^2x)=2+\cos x\Rightarrow\]\[0=\cos x(2\cos x+1)\]
We have,
\[\cos x=0\Rightarrow x=\frac{\pi}{2}\]\[2\cos x+1=0\Rightarrow x=2\pi/3\]

Answer 5

But how do you know if its the general solution or the specific

Answer 6

You can only obtain a specific solution, because there is a limited range in the problem. So I must correct my first answer in the first problem, k must be zero there, because there are only these two solutions.

In the second question the range specified is shorter, but again, is a specific solution.

Answer 7

How would we do the third one?
tan(3x)(tanx-1)=0

Answer 8

I don't understand how to work these out.

Answer 9

This is a product of two factor equal to zero, so
\[\tan(3x)=0\]
gives you a solution. In this case, in order to get zero in the tangent, the argument must be zero, so,
\[3x=0\Rightarrow x=0+2\pi k\]
For the other solution we have,
\[\tan x-1=0\Rightarrow \tan x=1\]
In order to get 1 in the tangent, x must be,
\[x_1=\pi/4+2\pi k\]
But, there is another angle whose tangent is 1, in the third quadrant,
\[x_2=5\pi/4+2\pi k\]
These two solutions can be put together in one,
\[x=\pi/4+\pi k\]

Answer 10

Awesome!!!!! Your great. Thanks for doing it step by step so I know what I'm actually doing. I hate when people just give you the answer. I like to learn how to do it

Answer 11

No problem ;)