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Quadratic trigonometric equations help? Solve x for 2sinsecx - 2sqrt(3)sinx = 0
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0<= x <= 2pi
The solutions i got were 0.955, and 5.328 The back of my book says the same thing except it also says 0 and 2pi
Here is how i did it: 2sinsecx = 2sqrt(3)sinx divide both sides by 2sinx secx = sqrt(3) cosx = 1/sqrt(3)
You did good, but you have to account for the part where the LHS is simply 0. So, sinx = 0 is also a solution.
i see what i did wrong, but why do i have to account for sinx = 0? I thought i already cancelled it out when i divided?
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You should not have cancelled out. Just factored.
|dw:1387148508328:dw|
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