sin⁡θ+3cos⁡θ=6x−x2−11,0≤θ≤4π,x∈R holds for

sin⁡θ+3cos⁡θ=−2−x2−6x+9=−2−(x−3)2∴sin⁡θ+3cos⁡θ≥−2 and −2−(x−3)2≤−2As a result, we have sin⁡θ+3cos⁡θ=−2 and then x = 3∴x=3 and cos⁡θ−π6=−1, i.e., θ−π6=π,3π

sin⁡θ+3cos⁡θ=6x−x2−11,0≤θ≤4π,x∈R holds for

sin⁡θ+3cos⁡θ=−2−x2−6x+9=−2−(x−3)2∴sin⁡θ+3cos⁡θ≥−2 and −2−(x−3)2≤−2As a result, we have sin⁡θ+3cos⁡θ=−2 and then x = 3∴x=3 and cos⁡θ−π6=−1, i.e., θ−π6=π,3π

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$\sin θ + \sqrt{3} \cos θ = 6 x - x^{2} - 11, 0 \leq θ \leq 4 π, x \in R$ holds for

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values of x and $θ$

one value of x and two values of $θ$

two values of x and two values of $θ$

two point of values of $(x, θ)$

answer is B.

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Detailed Solution

$\begin{aligned} \sin θ + \sqrt{3} \cos θ = - 2 - (x^{2} - 6 x + 9) = - 2 - (x - 3)^{2} \\ ∴ \sin θ + \sqrt{3} \cos θ \geq - 2 and - 2 - (x - 3)^{2} \leq - 2 \end{aligned}$
As a result, we have $\sin θ + \sqrt{3} \cos θ = - 2$ and then x = 3
$∴ x = 3 and \cos (θ - \frac{π}{6}) = - 1, i.e., θ - \frac{π}{6} = π, 3 π$