Question 48064: "solve the following trigonometric eqaution for the given domain (cont.)":
can someone show me how to solve the below eqautions and explain them to me? thanks.
i dont understand what you have to do with the "x" and a lot of other things. please help thank you.
a.) sin 3x =  ....for all x
b.) tan 2x= 0 ....on [ , ]
c.) sec x= 2......for all x
d.) sin^2 x + sin x=0....on [0,2]
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! a.) sin 3x = (sqrt3)/2 ....for all x
The reference angle that has this sin value is pi/3
The sin is positive in the 1st and 2nd Quadrants.
So, looking at all x, 3x= pi/3+2npi or 3x=(2/3)x+2npi
Then x=pi/9 + (2/3)npi or x= (pi/9)x+(2/3)npi
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tan 2x= 0 ....on [pi/2,3pi/2]
The reference angle whose tan value is 0 is 0 and pi
So, looking at the given interval 2x=pi and x=pi/2
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c.) secx= 2......for all x
The reference angle for sec value of 2 is pi/3
The sec is positive in the 1st and 4th quadrants.
so, looking at all x, x=pi/3 + 2npi or x=(5/3)pi + 2npi
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d.) sin^2 x + sin x=0....on [0,2pi]
sinx(sinx+1)=0
sinx=0 or sinx=-1
sinx=0 on the interval when x=0, or x=pi, or x=2pi
sinx=-1 on the interval when x=(3/2)pi
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Cheers,
Stan H.
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