# Solve this equation: 0 ≤ x ≤ 2pi: 2sin²x - 2sinx - 4 = 0?

SOLUTION: Solve the equations of [0, 2pi]: cosx(2sinx-1)=0 and (cos^2)x=(sin^2)x ... (cos^2)x-(sin^2)x=0 cos2x=0 2x=π/2, 3π/2 x=π/4, ... - Read more

Solve the equation: cos²(x) ... 2 = 0 cos(x) = (1 +/- sqrt(1 + 8)) / 2 cos(x) = (1 +/- 3) / 2 cos(x) = 4/2 , -2/2 cos(x) = 2 , ... [0 , 2Pi). 2sin (x) cos (-x) = 2 ... - Read more

## Solve this equation: 0 ≤ x ≤ 2pi: 2sin²x - 2sinx - 4 = 0? resources

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Results 1 to 4 of 4 Thread: Solve the equation for x if 0 ≤ x < 2pi: 2 cos(x) ... Solve the equation for x if 0 ≤ x < 2pi: 2 cos(x) + 3 tan(x) = 3 sec(x)

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... [ 0 , 2 pi ] Sec ^ 2 x - Tan x = 1 -----> ... Hence the solution will be x = 0 , ∏ , ∏ / 4 or 5∏ / 4. ... solve equation usnig domain (0,2pi) ...

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### Solve the equation sin2x-sin4x=0? - Homework Help - eNotes.com

sin2x-sin4x=0. sinx-siny = 2cos(x+y)/2*sin(x-y)/2 ==> sin2x-sin4x= 2cos(3x)*sin(-x) = 2cos3x*-sinx= -2cos3xsinx=0 ==> -2cos3x*sinx= 0. then cos3x=0 OR sinx=0

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If someone could point me in the right direction of where to go with these sorts of problems, I would be so greatful :p Solve the equation for x, where 0

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