# Math Help - Trigonometric Equations

1. ## Trigonometric 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. 2sin2= 2 + cosx
3. 2sin2x + 3sinx + 1 = 0

Thank you so much!!!!!!!!!!!!!!!!!!!!
Isabel

2. Originally Posted by cuteisa89
Can somebody help me with these?

Solve the equation:
1. tan3x (tanx - 1)=0
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

Find all solutions of the equation in the interval [0, 2π)
2. 2sin2= 2 + cosx
3. 2sin2x + 3sinx + 1 = 0
are these 2's supposed to be squares?

3. I'll assume the bold 2's mean squares, since i got no response
Originally Posted by cuteisa89
Find all solutions of the equation in the interval [0, 2π)
2. 2sin2= 2 + cosx
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]

3. 2sin2x + 3sinx + 1 = 0
2sin^2x + 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]

4. 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

5. Originally Posted by cuteisa89
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
well as you can see, i already solved it with 2 as an exponent

it's fine as long as you understand it

6. Originally Posted by Jhevon
tan3x (tanx - 1) = 0
=> tan3x = 0 or tanx - 1 = 0

=> 3x = arctan0
=> x = (1/3)arctan0
=> x = k*pi for k an integer
this last line should be:

=> x = (k/3) 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?
RonL

7. Originally Posted by CaptainBlack
this last line should be:

=> x = (k/3) pi for k an integer

RonL
right, i forgot about the 1/3