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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