Can somebody help me with these?
Solve the equation:
1. tan3x (tanx - 1)=0
Find all solutions of the equation in the interval [0, 2π)
2. 2sin2= 2 + cosx
3. 2sin2x + 3sinx + 1 = 0
Thank you so much!!!!!!!!!!!!!!!!!!!!
Isabel
tan3x (tanx - 1) = 0
=> tan3x = 0 or tanx - 1 = 0
=> 3x = arctan0
=> x = (1/3)arctan0
=> x = k*pi for k an integer
=> tanx = 1
=> x = arctan(1)
=> x = pi/4 + k*pi for k an integer
are these 2's supposed to be squares?
Find all solutions of the equation in the interval [0, 2π)
2. 2sin2= 2 + cosx
3. 2sin2x + 3sinx + 1 = 0
I'll assume the bold 2's mean squares, since i got no response
2sin^2x = 2 + cosx
=> 2(1 - cos^2x) = 2 + cosx
=> 2 - 2cos^2x = 2 + cosx
=> 2cos^2x + cosx = 0
=> cosx(2cosx + 1) = 0
=> cosx = 0 or 2cosx + 1 = 0
cosx = 0
=> x = arcos0
=> x = pi/2 , 3pi/2 for x in [0, 2pi]
2cosx + 1 = 0
=> x = arccos(-1/2)
=> x = 2pi/3 , 4pi/3 for x in [0, 2pi]
2sin^2x + 3sinx + 1 = 03. 2sin2x + 3sinx + 1 = 0
this is just a quadratic equation cleverly disguised. You can see this if you replaced sinx with y for example, you would end up with 2y^2 + 3y + 1 = 0 so we will treat this as a quadratic and try to foil.
=> (2sinx + 1)(sinx + 1) = 0
=> 2sinx + 1 = 0 or sinx + 1 = 0
2sinx + 1 = 0
=> x = arcsin(-1/2)
=> x = 7pi/6, 11pi/6 for x in [0, 2pi]
sinx + 1 = 0
=> x = arcsin(-1)
=> x = 3pi/2 for x in [0, 2pi]