Asked by brittney
1.Solve, finding all solutions in [0, 2π).
cosx sin2x + sinx cosx - sinx = 0
2.Solve, finding all solutions in [0, 2π).
3sec x = 2tan^2x
cosx sin2x + sinx cosx - sinx = 0
2.Solve, finding all solutions in [0, 2π).
3sec x = 2tan^2x
Answers
Answered by
Steve
cosx sin2x + sinx cosx - sinx = 0
cosx (2sinxcosx) + sinxcosx - sinx = 0
sinx(2cos^2x + cosx - 1) = 0
sinx(2cosx-1)(cosx+1) = 0
now it's easy, right?
3secx = 2tan^2x
3secx = 2(sec^2x-1)
2sec^2x - 3secx - 2 = 0
(2secx+1)(secx-2)
...
cosx (2sinxcosx) + sinxcosx - sinx = 0
sinx(2cos^2x + cosx - 1) = 0
sinx(2cosx-1)(cosx+1) = 0
now it's easy, right?
3secx = 2tan^2x
3secx = 2(sec^2x-1)
2sec^2x - 3secx - 2 = 0
(2secx+1)(secx-2)
...
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