Trigonometry/solving equations

[whitmack]

I need some help solving the equations below with the interval being [0, 2pie)

sin 2x + sinx =0 I have come up with 0, pie being the solutions, are there any more? are 2pie/3 and 4pie/3 solutions too?


sin^2 2x=1 I have come up pie/4, 3pie/4, 5pie/4,7pie/4 as solutions, is this right?


2+ 13sin x=14 cos^2x This one i have no clue...help please

 

[skeeter]

sin(2x) + sinx = 0
2sinxcosx + sinx = 0
sinx(2cosx + 1) = 0

sinx = 0 ... x = 0, x = pi
cosx = -1/2 ... x = 2pi/3, x = 4pi/3

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2 + 13sinx = 14cos^2x
2 + 13sinx = 14(1 - sin^2x)
2 + 13sinx = 14 - 14sin^2x
14sin^2x + 13sinx - 12 = 0
(7sinx - 4)(2sinx + 3) = 0

sinx = 4/7 ... x = arcsin(4/7) ... two solutions, one in quad I and one in quad II, get out your calculator

sinx = -3/2 ... no solutions for x.

 

[whitmack]

Thank you skeeter, I appreciated the help!


WRM