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i give medals! please help!
you can factor the common term of sin x sin² x + sin x = 0 (sin x)(sin x + 1) = 0 Now set both factors equal to 0 and solve these two equations sin x = 0 sin x + 1 = 0
I am very confused
i only have a few more minutes to answer this please help and explain!
So you have, \[\sin^2(x) + \sin(x) = 0\] By the definition of a exponent \[x^2 = x*x\] etc so, \[\sin^2(x) + \sin(x) = 0\] can be written as: \[\sin(x)\sin(x) + \sin(x) = 0\] Now you can factor out sin(x): \[\sin(x)(\sin(x) + 1) = 0\] Now look at both terms sin(x) and (1 + sin(x)) When one of them equals 0 you have a solution
sin² x and sin x have a common term of sin x so it can be factored out giving the equation as (sin x)(sin x + 1) = 0 If two thing multiply to be 0, one or both of them is 0. so we can write thes the two equations sin x = 0 and sin x + 1 = 0 To solve sin x = 0, look on the unit circle and pick out all the angle where sine is 0. To solve sin x + 1 = 0, subtract the 1 from both sides sin x = -1 now pick out the angles that have a sine of -1.
so for (1 + sin(x)) = 0 sin(x) must equal - 1 what value of x makes sin(x) = -1? For the second term sin(x) = 0 what value of x makes sin(x) equal 0 Look at the graph of sin(x) http://www.wolframalpha.com/input/?i=graph+of+sin%28x%29+between+0+and+2pi
Remember that you are only looking for solutions for the interval [0,2pi)
is there answer x= 0,pi,4pi/3,5pi/3?
0 and pi are solutions to sin x = 0. the other two aren't solutions to either equation. Where is sine equal to -1 between 0 and 2pi?
when x = 0 sin(0) = 0 so x =0 is a solution sin(4pi/3) does not equal 0 or -1 so it is not a solution sin(5pi/3) does not equal 0 or -1 so it is not a solution sin(pi) = 0 so it is a solution
0, pi, 3pi/2
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