how do i solve for all angles [0,2pi]????

sin2x + sin x = 0

cos^2-cos 2x = 0

PLEASE HELP!!! THANKS

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- July 12th 2005, 04:15 PMJustardmsSolve for all angles [0,2pi] HELP!! PLEASE
how do i solve for all angles [0,2pi]????

sin2x + sin x = 0

cos^2-cos 2x = 0

PLEASE HELP!!! THANKS - July 13th 2005, 12:09 AMticbol
You expand the equations in terms of angle x only. Meaning, expand sin(2x) and cos(2x) into trig functions/function of x only. The idea to see if you can form equations in x only that you can solve algebraocally.

[Say what? :-) . Okay, it is really easier to explain with numbers...]

sin(2x) +sin(x) = 0

2sin(x)cos(x) +sin(x) = 0

[sin(x)]*[2cos(x) +1] = 0

sin(x) = 0

x = arcsin(0)

x = 0, pi, and 2pi ---***

2cos(x) +1 = 0

2cos(x) = -1

cos(x) = -1/2

x = arccos(-1/2)

x is in the 2nd and 3rd quadrant, arccos(1/2) is 60 degrees or pi/3, so,

In the 2nd quadrant, x = pi -pi/3 = 2pi/3 ---***

In the 3rd quadrant, x = pi +pi/3 = 4pi/3 ----***

Therefore, in the interval [0,2pi],

x = 0, 2pi/3, pi, 4pi/3, 2pi --------------answer.

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cos^2(x) -cos(2x) = 0

cos^2(x) -[cos^2(x) -sin^2(x)] = 0

sin^2(x) = 0

sin(x) = 0

x = arcsin(0)

x = 0, pi, 2pi ----answer.

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To see if those answers are correct, try plugging them into the original equations. If they make the original equations true, then the answers are correct.

[Hint: They are correct.] - July 13th 2005, 05:56 AMJustardmsThanks
Thank You Very Much!!!