finding values of x in the interval {0 degrees, 360 degrees)

Mathematics
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(2sin*squared*x-1)((sin*squared*x-1)=0

May 12th, 2015

so we have: (2*sin^2(x)-1)*(sin^2(x)-1) = 0. This means that one of the two factors must be equal to zero. So let's start with the first factor: 2*sin^2(x) - 1 = 0. This means that sin^2(x) = 1/2, so sin (x) = +/- sqrt(2)/2. Next this needs to be memorized: for what angle is sin(x) = sqrt(2)/2. This is the angle 45 degrees. However there are more angles. Notice that because we don't care about the signs and because the value isn't +/- 1 which is "in between quadrants," there will be 4 solutions to sin(x) = +/- sqrt(2)/2, but all of them will make a 45 degree angle with the x axis. 

So we have x = 45, 180-45, 180+45, and 360-45. After simplifying we get: x = 45, 135, 225, 315. Alternatively, the second factor could equal zero yielding: sin^2(x) = 1 --> sin = +/- 1. This will occur when x = 90, and when x = 270. Note that sin(90) = 1 is another memorized value, we get that the second value also must make a 90 degree angle with the x axis. All together the solutions are x = 45, 90, 135, 225, 270, 315.

May 12th, 2015

Please ask if anything is unclear.


May 12th, 2015

how did you get sqrt 2/2 from sin^2(x)= 1/2

May 12th, 2015

Okay so we take the square root of both sides. sin(x) = sqrt(1/2) = sqrt(1)/sqrt(2) = 1/sqrt(2). Next, we rationalize the fraction by multiplying by sqrt(2)/sqrt(2). 1/sqrt(2)*sqrt(2)/sqrt(2) = (1*sqrt(2))/(sqrt(2)*sqrt(2)) = sqrt(2)/2

May 12th, 2015

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