Expert: Sherman D. - 3/29/2006

Question
I have two.  I hope that's okay.  The section is solving trig equations.
(1) 2sin(30 + x)=3cosx  (The 30 is 30 degrees. solve each equation for 0<=x<=360 degrees)

(2) Cos^-1(2x)=Sin^-1(x) (solve each equation for 0<=x<=2pi)

Thank you.

Answer
Unless i am doing something wrong, i can't find a match to either.

When its working, go to www.quickmath.com, click on Solve under Equations, type in 2sin(30 + x) = 3cos(x), then click enter and it will solve for you.

1.)
2sin(30 + x) = 3cos(x)

(cos(x))/(sin(30 + x)) = (2/3)

y = cos(x)
((0,360),1),((60,300),(1/2)),((90,270),0),((120,240),(-1/2)),(180,-1)

y = sin(30 + x)
((180,300),(-1/2))((0,120),(1/2)),(60,1),(120,(1/2)),(240,-1),(330,0)

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2.) On this problem, you can't have anything higher than 1 or lower than -1, so what do you mean by 0 <= x <= 2pi.
cos^-1(2x) = sin^-1(x)
y = sin^-1(x)
(-1,(3pi/2)),((-1/2),((11pi/6) or (7pi/6))),(0,0),((1/2),((pi/6) or (5pi/6))),(1,(pi/2))

For this one, you can't have anything higher than .5 or lower than -.5, because that would make this problem undefined.
y = cos^-1(2x)
((-1/2),pi),((-1/4),(2pi/3)),(0,((pi/2) or (3pi/2))),((1/4),(pi/3)),((1/2),((0 or 2pi))

If there was no ^-1, i could tell you the answers.