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

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- May 5th 2007, 10:09 PMcuteisa89Trigonometric Equations
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. 2sin**2**= 2 + cosx

3. 2sin**2**x + 3sinx + 1 = 0

Thank you so much!!!!!!!!!!!!!!!!!!!! :)

Isabel - May 5th 2007, 10:20 PMJhevon
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

Quote:

Find all solutions of the equation in the interval [0, 2π)

2. 2sin**2**= 2 + cosx

3. 2sin**2**x + 3sinx + 1 = 0

- May 5th 2007, 10:50 PMJhevon
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]

Quote:

3. 2sin**2**x + 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] - May 6th 2007, 03:27 PMcuteisa89
Sorry I meant the 2 as a exponent but it didn't come out quite the right way. Thank you for all the help though... It is much appreciate it! :)

Isabel - May 6th 2007, 03:28 PMJhevon
- May 6th 2007, 09:30 PMCaptainBlack
- May 6th 2007, 09:32 PMJhevon