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cheahjimmy
@cheahjimmy
September 2022
1
11
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How many solutions does the equation 3sin²xcosx = cosx
in the interval 0 ≤ x < 2π have
please show your work
Answers & Comments
SugoiOverload
3sin²xcosx = cosxsinx
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}
2 votes
Thanks 6
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Answers & Comments
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}