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How many solutions does the equation 3sin²xcosx = cosx

in the interval 0 ≤ x < 2π have

please show your work

1
by cheahjimmy

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in the interval 0 ≤ x < 2π have

please show your work

by cheahjimmy

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3sin²xcosx - sinxcosx = 0

sinxcosx(3sinx - 1) = 0

(sinx)(cosx)(3sinx - 1) = 0

sinx = 0

x = arcsin(0)

x = 0 but since the problem asks for all x between 0 and 2π that satisfies the equation, then x = 0 or x= π

cosx = 0

x = arccos(0)

x = π/2 but since the problem asks for all x between 0 and 2π that satisfies the equation, then x = π/2 or x= 3π/2

3sinx - 1 = 0

3sinx = 1

sinx = 1/3

x = arcsin(1/3)

x = 0.3398 I just used a calculator to get that, so well...

Combining all the x's that we get,

SS: {0, π/2, 3π/2, π, 0.3398}