Question #26480

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Answer 1 · 2018-02-22T08:11:42

$θ = -π/18$ or $θ= -π/6$

$sin(3θ−π/6)=−(√3)/2$

the identity is as follows
sin(A - B) = sin A cos B - cos A sin B

$[Sin 3θ][Cos(π/6)] - [Cos 3θ][Sin(π/6)] = −(sqrt3)/2$

using special triangles,

$Cos(π/6) = (√3)/2$ , $Sin(π/6) = 1/2$

Thus,

$[(sqrt3)/2][Sin 3θ] - (1/2)[Cos 3θ] = −(sqrt3)/2$

multiply both sides of the equation by 2

$(sqrt3)[Sin 3θ] - [Cos 3θ] = -sqrt3$

then $cos 3θ = sqrt(1- sin^2 3θ)$

$(sqrt3)[Sin 3θ] - sqrt(1- sin^2 3θ) = -sqrt3$

$(√3)[Sin 3θ] +sqrt3 =sqrt(1- sin^2 3θ)$

squaring both sides and simplifying

$3sin^2 3θ +6[Sin 3θ] +3 = 1- Sin^2 3θ$

transposing all the terms from the right side to the left side and then combining like terms. Simplifying

$2sin^2 3θ + 3[Sin 3θ] +1 = 0$

then let $Sin 3θ = x$

$2 x^2 + 3x +1 = 0$
then factor

$(2x +1) ( x+1) = 0$

then $x= -(1/2)$ and $x = -1$

but $Sin 3θ = x$

solving for $θ$

$θ = -π/18$ or $θ= -π/6$