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# Find all the solutions and indicate solutions in interval [0,2π)?

sin2x - sinx cosx = cosx

i need to see the work, that would help a lot.

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- Adrian SLv 78 years ago
Ok Greg--

Using the double angle identity for sin(2x) you'll get 2sin(x)cos(x)

Now subtract and bring over the cos(x)and set = 0 and you'll get

cos(x)*(sin(x) -1) = 0.

By ZPP cos(x) = 0 when x = pi/2 and 3pi/2

and sin(x) = 1 when x = pi/2

Source(s): Math tutor for several years. - KateLv 48 years ago
sin 2x - sin x cos x = cos x

Substitute in sin2x = 2sinxcosx [from sin(a + b) = sin a cos b + cos a sin b]:

2sin x cos x - sin x cos x = cos x

sin x cos x - cos x = 0

Factor out cos x:

cos x(sin x - 1) = 0

cos x = 0

x = π/2, 3π/2

sin x = 1

x = π/2

∴ x = π/2, 3π/2

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