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# Final exam:(

-7
104
2
+318

Solve each trigonmetric equation in the interval [0, 2pi) by first squarring both sides.

sq rt(3) cos x=sin x+1

solution set??????

type an exact answer, using pi as needed.

May 17, 2020
edited by mharrigan920  May 17, 2020

#1
+21953
+2

Solve by squaring both sides:

sqrt(3)·cos(x)  =  sin(x) + 1

sqrt(3)·cos(x) ]2  =  [ sin(x) + 1 ]2

3·cos2(x)  =  sin2(x) + 2·sin(x) + 1

Since  cos2(x)  =  1 - sin2(x)

3·( 1 - sin2(x) )  =  sin2(x) + 2·sin(x) + 1

3 - 3·sin2(x)  =  sin2(x) + 2·sin(x) + 1

0  =  4sin2(x) + 2sin(x) - 2

0  =  2sin2(x) + sin(x) - 1

0  =  ( 2sin(x) - 1 )( sin(x) + 1 )

So:  either       2sin(x) - 1  =  0       or     sin(x) + 1  =  0

2sin(x)  =  1                     sin(x)  =  -1

sin(x)  =  ½                          x  =  sin-1(1)

x  =  sin-1( ½  )

Now, you'll need to finish this ...

May 17, 2020
#2
+110735
+2

mharrigen920,

Why did you vote geno down?

I suppose this is your standard response when people invest their time and expertise in helping you.

May 17, 2020