Asked by Sean
Solve 2cos^2(x) + 4sin(x)-3 = 0, giving all solutions in the interval 0 ≤ x ≤ 2π
Answers
Answered by
Damon
I already did this today
180-30
360 - 30
270 in degrees
multiply degrees by pi/180 to get radians if that is what you need
180-30
360 - 30
270 in degrees
multiply degrees by pi/180 to get radians if that is what you need
Answered by
Damon
http://www.jiskha.com/display.cgi?id=1447870446
Answered by
Damon
oh, you changed a number
2 (1 -sin^2) + 4 sin - 3 = 0
2 - 2 sin^2 + 4 sin - 3 = 0
-2 sin^2 + 4 sin - 1 = 0
2 sin^2 - 4 sin + 1 = 0
(2 sin -1)(2 sin-1) = 0
sin = 1/2
angle = 30 or 150 degrees
2 (1 -sin^2) + 4 sin - 3 = 0
2 - 2 sin^2 + 4 sin - 3 = 0
-2 sin^2 + 4 sin - 1 = 0
2 sin^2 - 4 sin + 1 = 0
(2 sin -1)(2 sin-1) = 0
sin = 1/2
angle = 30 or 150 degrees
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