Asked by Josh
                Getting ready for trig semester finals. Can't figure out how they are simplifying these equations. (these are the review not actual test)
tanx-sqr3=2tanx
and
2sin^2+sinx=0
            
        tanx-sqr3=2tanx
and
2sin^2+sinx=0
Answers
                    Answered by
            Reiny
            
    tanx-sqr3=2tanx 
tanx = -√3
tanx is negative , so x must be in quads II or IV
I know tan 60° = √3
so x = 180-60= 120°
or
x = 360-60 = 300°
in radians that would be 2π/3 or 5π/3
2sin^2 x + sinx = 0
sinx(2sinx + 1) = 0
sinx = 0 or sinx = -1/2
sinx = 0 ----> x = 0, 180° or 360°
or sinx = -1/2
since sinx is negative, x must be in III or IV
I know sin 30 = +1/2
x = 180+30 = 210 or x = 360-30 = 330°
so x = 0,180, 360,210 , or 330
    
tanx = -√3
tanx is negative , so x must be in quads II or IV
I know tan 60° = √3
so x = 180-60= 120°
or
x = 360-60 = 300°
in radians that would be 2π/3 or 5π/3
2sin^2 x + sinx = 0
sinx(2sinx + 1) = 0
sinx = 0 or sinx = -1/2
sinx = 0 ----> x = 0, 180° or 360°
or sinx = -1/2
since sinx is negative, x must be in III or IV
I know sin 30 = +1/2
x = 180+30 = 210 or x = 360-30 = 330°
so x = 0,180, 360,210 , or 330
                    Answered by
            Josh
            
    I get the second one, but the first I don't. What identity are you using for the first to get from line one to two?
    
                    Answered by
            Steve
            
    just rearranging terms, collecting tan's on the left, numbers on the right. Pretty mysterious, eh?
    
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