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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
tan3x (tanx - 1) = 0Originally Posted by cuteisa89
=> 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]
well as you can see, i already solved it with 2 as an exponentSorry 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
it's fine as long as you understand it
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