#12 5.05 Precalculus questions

Mathematics
Tutor: None Selected Time limit: 1 Day

May 15th, 2015

Best Answer:  sin2x sinx - cosx=0
--> [2sin(x)cos(x)]sin(x) - cos(x) = 0
--> 2sin^2(x)cos(x) - cos(x) = 0
--> 2[1 - cos^2(x)]cos(x) - cos(x) = 0
--> 2cos(x) - 2cos^3(x) - cos(x) = 0
--> 2cos^3(x) + cos(x) = 0
Let y = cos(x), then
2y^3 + y = 0
--> y(2y^2 + 1) = 0
--> y = 0
--> cos(x) = 0
i.e. x = (2k+1)pi/2 for k = ..., -1, 0, 1, ....

May 15th, 2015

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