Trig equation: Solve sin x = - (radical 2) - cos x on 0, 2pi

[as8906]

I have no idea where to start, I tried a few things that were apparently wrong, so any help would be appreciated.
I couldn't get the proper symbols, but the substitutions should be obvious.

Just to avoid confusion, "<=" means "less than or equal to"

solve the equation on the interval 0 <= x < 2pi

sin x = - (radical 2) - cos x

I have the answer to the problem, it's {5pi/4}, I just wanted to see how it's done. Thanks to whoever helps.

 

[chivox]

Re: Trig equation

I have no idea where to start, I tried a few things that were apparently wrong, so any help would be appreciated.
I couldn't get the proper symbols, but the substitutions should be obvious.

solve the equation on the interval 0 <= x < 2pi
sin x = - (radical 2) - cos x
I have the answer to the problem, it's {5pi/4}, I just wanted to see how it's done. Thanks to whoever helps.

Could you write down what steps you took? Probably we can see where they start going wrong. Based on my experience, the solution will come a lot quicker if you show us at least the first step. (I can think of many things you might do to approach the solution, so let's see what your thoughts were.)

 

[galactus]

\(\displaystyle sin(x)+cos(x)=-\sqrt{2}\)

Rewrite as \(\displaystyle \sqrt{2}sin(x+\frac{\pi}{4})=-\sqrt{2}\)

\(\displaystyle sin(x+\frac{\pi}{4})=-1\)

Finish?.

 

[as8906]

Wait, how did you do that? Could you explain the steps and what properties allowed you to do that?

I also added cos x to both sides, and ended up with: sin x + cos x = -radical2. After that I squared both sides, but I got some other answer. Could you tell me what I did wrong?

 

[mmm4444bot]

I also added cos x to both sides, and ended up with: sin x + cos x = -radical2. After that I squared both sides, but I got some other answer. Could you tell me what I did wrong?

Without seeing your work, I cannot see where you went wrong.

After squaring the equation, did you simplify to sin(x) * cos(x) = 1/2?

The Quadrant I solution is one of the standard angles for which you should have both the sine and cosine memorized. Using that as a reference angle, find the second solution.

When you square an equation during solving, you need to check your result in the original equation because squaring can introduce false results.

In other words, only one of the two solutions above satisfies the original equation.

 

[as8906]

yes, i ended up with sinx cosx = 1/2. I guess just from looking at this, pi/4 would work as an answer, because:

sin pi/4 = 1/radical 2

cos pi/4 = 1/radical 2

1/radical 2 times itself would be 1/2. But how do I know that the answer is not pi/4 and instead is 5pi/4? Should I plug the values into the original equation?

 

[mmm4444bot]

... But how do I know that the answer is not pi/4 and instead is 5pi/4?

Should I plug the values into the original equation?

Did I already answer these questions? (This is a rhetorical question.)

... you need to check your result in the original equation because squaring can introduce false results.

In other words, only one of the two solutions above satisfies the original equation.

 

[as8906]

ok thanks