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in Trigonometry by (94.7k points)
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The equation `x^3=3/4x=-(sqrt(3))/8` is satisfied by `x=cos((5pi)/(18))` (b) `x=cos((7pi)/(18))` `x=cos((23pi)/(18))` (d) `x=cos((17pi)/(18))`
A. `x=cos((5pi)/(18))`
B. `x =cos((7pi)/(18))`
C. `x=cos((23pi)/(18))`
D. `x=-sin((7pi)/(9))`

1 Answer

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Best answer
Correct Answer - A::B
Let `x = cos theta`
`rArr 4 cos^(3) theta -3 cos. theta =-(sqrt(3))/(2)`
`rArr cos 3 theta = cos.(5pi)/(6)`
`rArr 3 theta = 2n pi pm (5pi)/(6), n in Z`
`rArr theta =(2n pi)/(3)pm(5pi)/(18),, n in Z`
Put `n=0, theta =(5pi)/(18)`
`n=1, theta =(2pi)/(3)+(5pi)/(18)=(17pi)/(18)`
`theta =(2pi)/(3)-(5pi)/(18)=(7pi)/(18)`

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