Solving 3 cos x + 4 = 0 cos x = -4/3

[cscott]

3 cos x + 4 = 0
cos x = -4/3

How do I solve this? I can't take the arcsine...

---

sin (3x - 40) = 0

Is the general solution 73.3 + 120k | k E I?

 

[robphy]

3 cos x + 4 = 0
cos x = -4/3

How do I solve this? I can't take the arcsine...

Did you try to use complex numbers or hyperbolic trig functions?

 

[hotvette]

How do I solve this? I can't take the arcsine...

Could try graphing it and see where it crosses zero.

 

[Jameson]

It's not going to cross the x-axis. You'll have to encorporte complex numbers like robphy said.

 

[cscott]

We don't cover complex numbers until the end of the year. Does this mean I say it has no solution?

 

[JoshHolloway]

Why can you not take the cosine inverse of x?

 

[cscott]

Why can you not take the cosine inverse of x?

4/3 > 1 so you get a domain error... or at least that's how I see it.

 

[benjamincarson]

3 cos x + 4 = 0
cos x = -4/3

Unless I'm missing something, your given value for cos x is incorrect. Any value of the cosine function is in a way "stuck" between -1 and 1.

Use a calculator and take the cosine of any degree/radian measure you wish, it will always be between -1 and 1.

 

[cscott]

Unless I'm missing something, you're given value for cos x is incorrect. Any values of the cosine function is in a way "stuck" between -1 and 1.

Use a calculator and take the cosine of any degree/radian measure you wish, it will always be between -1 and 1.

I got -4/3 from rearranging the equation. The fact that 4/3 > 1 is my problem.

 

[benjamincarson]

Is this an equation that needs solving, or an identity that needs verifying?

 

[cscott]

"Solve for all possible values of x."

 

[benjamincarson]

no solution

 

[cscott]

Thank you.

 

[cscott]

I have another problem

sin^2 x = 3/4

for one of the possible answers I get \frac{\pi}{3} + 2\pi k but my book says it should be \pi instead of 2\pi for the period. How come?

 

[whozum]

The sin value is squared, so the negative values square out to give a solution too.

 

[cscott]

The sin value is squared, so the negative values square out to give a solution too.

Yes I know. Let me clarify: \frac{\pi}{3} + 2\pi k is one possible solution but my book says it should be \frac{\pi}{3} + \pi k and I don't know why.

 

[robphy]

e^{i\pi}=-1

 

[cscott]

e^{i\pi}=-1

I haven't done complex numbers so I don't really know the significance of that expression.

 

[TD]

I have another problem

sin^2 x = 3/4

for one of the possible answers I get \frac{\pi}{3} + 2\pi k but my book says it should be \pi instead of 2\pi for the period. How come?

\sin ^2 x = \frac{3}<br /> {4} \Leftrightarrow \sin x = \pm \sqrt {\frac{3}<br /> {4}} = \pm \frac{{\sqrt 3 }}<br /> {2}

Now determine the solution in both cases (the + and the - case).

 

[cscott]

\sin ^2 x = \frac{3}<br /> {4} \Leftrightarrow \sin x = \pm \sqrt {\frac{3}<br /> {4}} = \pm \frac{{\sqrt 3 }}<br /> {2}

Now determine the solution in both cases (the + and the - case).

I would say

\frac{\pi}{3} + 2\pi k | k \epsilon I, \frac{4\pi}{3} + 2\pi k | k \epsilon I

but my book says

\frac{\pi}{3} + \pi k | k \epsilon I, \frac{4\pi}{3} + \pi k | k \epsilon I

and I don't know why

 

[TD]

Also remember the supplementary solution, \sin \left( \alpha \right) = \sin \left( {\pi - \alpha } \right). You immediately have this if you don't multiply with 2k*pi but just with k*pi.

 

[cscott]

Also remember the supplementary solution, \sin \left( \alpha \right) = \sin \left( {\pi - \alpha } \right). You immediately have this if you don't multiply with 2k*pi but just with k*pi.

Ohhh, I see. Thanks a lot!

By the way, how do you read the <=> arrow you used and how is it different than just =>?

 

[TD]

No problem!

 

[Diane_]

The "=>" is an implication arrow - it means that, if the left-hand side is true, the right-hand side is also true.

The "<=>" is basically an implication arrow going both ways, and is generally read "if and only if". It's a much more powerful condition, meaning that if either side is true, the other side is also true. Outside of definitions and relatively trivial things, you don't get one of those very often. When you do, it's worth paying attention to.

 

[cscott]

Wait! Let me just check my understanding: the solution to the equation I posted can be either the one including pi/3 or the one inlcluding 4pi/3 and each solution has to encorporate the negative and positive sqrt(3)/2?

I say this because pi/3 + pi gives -sqrt(3)/2 but I got the pi/3 in the first place using sin x = +sqrt(3)/2

Is this correct? It doesn't seem like it...

 

[TD]

Wait! Let me just check my understanding: the solution to the equation I posted can be either the one including pi/3 or the one inlcluding 4pi/3 and each solution has to encorporate the negative and positive sqrt(3)/2?

I say this because pi/3 + pi gives -sqrt(3)/2 but I got the pi/3 in the first place using sin x = +sqrt(3)/2

Is this correct? It doesn't seem like it...

From the positive root, the solution pi/3 and its complementary solution pi-pi/3 = 2pi/3 follow.
From the negative root, the solution -pi/3 (or 5pi/3) and its complementary solution 4pi/3 follow.

You see?

 

[cscott]

From the positive root, the solution pi/3 and its complementary solution pi-pi/3 = 2pi/3 follow.
From the negative root, the solution -pi/3 (or 5pi/3) and its complementary solution 4pi/3 follow.

You see?

I understand that there is two solutions for each root but I don't see how that translates into the one solution per root. How do you get 2pi/3 from pi/3 + pi * k? Shouldn't I be able to put any integral value for "k" and get a positive sqrt(3)/2 answer in that case?

 

[TD]

For the positive root, we had: pi/3 and 2pi/3
For the negative root, we had: 4pi/3 and 5pi/3

Of course, all +2k*pi.

But now notice that 4pi/3 = pi/3 + pi and 5pi/3 = 2pi/3 + pi.
So you need to 'cross' the solutions of both roots, take them all together.
Then you can leave out the two redundant ones and use +k*pi instead of +2k*pi.

 

[cscott]

I get it! I get it! Thanks again for being patient with me.

It's like this was a one on one (fast posts)

 

[TD]

No problem, glad you understand