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Cosine equation solution set in an interval

Given the algebraic solution set for a cosine equation, find which solutions fall within an interval. Created by Sal Khan.

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• The interval (-pie/2,0),is arbitrary?. Could he have chosen any other interval?
• other people have asked the same problem check their answers and hopefully that will help you. I am asking myself the same question so I dont trust myself to answer
• , Sal encourages us to evaluate a portion of the equation on a calculator. Instead of coming up with approximately 0.22, I keep ending up at approximately 12.44. Can anyone explain why?
• 0.22 radians = 12.44°

Change the calculator mode to "RAD" instead of "DEG".
• Where did the interval [-π/2, 0] come from?
• That was just part of the given of the problem which puts our cos in the 3rd and 4th quadrant. So cos-1(x) will have positive values in the 1st and 4th quadrants and negative values in the 2nd and 3rd quadrants. Sal uses this limit to say we are only thinking about what is happening in the 3rd and 4th quadrant.
• Hi, why are we restricting the range of x to be in between [-pi/2, 0]? timestamp at approximately
• i think he chose arbitrary values for the range. the point of this video is just to teach us how to find values of x within a range
• I really don't understand why he choose that interval, what was the purpose of choosing it? since from the previous video he already graphed it a little past pi/2, but here he chose a invertval that does not even reach it?
• @ , why did he take -pi/2, was it just an arbitrary interval?
• Hi, why did SAL take [-Pi/2, 0] as the limit. Shouldn't the limit for the COS function be between 0 and Pi (0 and 3.14) ?

Considering the given value of -1/6 the coordinate turns out to be falling on second quadrant. But the same coordinate can be achieved on the vertically opposite side that is third quadrant. Since both these quadrants (second and third) give the stated value in the question (i.e. -1/6), I was assuming we should have restricted to one of these quadrant. Can someone help clarify this?
• The COS function (adj/hyp) on the unit circle is positive in the 1st and 4th quadrants, and negative in the 2nd and 3rd quadrants. If you went between the 1st and 2nd, the cos(first)=-cos(second), so you get two different answers.
• Why can't there be infinite x values that satisfy these situations? For example, if 0 & 1 satisfy it; why can't 0.67 satisfy it and every other decimal quantity in between 0 and 1?
• That's not how functions work. If two values satisfy a condition, everything in between needn't.

Think about this. Say I have x^2 = 4. Both x = -2 and x = 2 satisfy this. However, no number between -2 and 2 satisfies it.