The general value of x satisfying the equation \(\sqrt3\)sinx + cosx

1.	The general value of x satisfying the equation \(\sqrt3\)sinx + cosx = \(\sqrt3\) is given byA. x = nπ + (-1)n \(\frac{π}4\) + \(\frac{π}3\),n ∈ Z.B. x = nπ + (-1)n \(\frac{π}3\) + \(\frac{π}6\),n ∈ Z.C. x = nπ ± \(\frac{π}6\),n ∈ Z.D. x = nπ ± \(\frac{π}3\),n ∈ Z.
Answer» Cos²x = (√3 -√3sin x)² 1 - sin²x = 3 + 3sin²x - 6sin x 4sin²x - 6sin x+2 = 0 2sin²x - 3sin x+1 = 0 sin x = 1 or 0.5 We know, x = nπ + (-1)ⁿθ x = nπ + (-1)ⁿ\((\frac{π}2)\) or x = nπ + (-1)ⁿ\((\frac{π}6)\) Therefore, the values of x are \(\frac{π}6\),\(\frac{π}2\),\(\frac{5π}6\),π,\(\frac{13π}6\) So, these values are obtained for different value of n from the equation x = nπ + (-1)ⁿ\(\frac{π}2\) - \(\frac{π}6\),n ∈ Z. So, Option B

The general value of x satisfying the equation \(\sqrt3\)sinx + cosx = \(\sqrt3\) is given byA. x = nπ + (-1)n \(\frac{π}4\) + \(\frac{π}3\),n ∈ Z.B. x = nπ + (-1)n \(\frac{π}3\) + \(\frac{π}6\),n ∈ Z.C. x = nπ ± \(\frac{π}6\),n ∈ Z.D. x = nπ ± \(\frac{π}3\),n ∈ Z.

Answer»

Cos²x = (√3 -√3sin x)²

1 - sin²x = 3 + 3sin²x - 6sin x

4sin²x - 6sin x+2 = 0

2sin²x - 3sin x+1 = 0

sin x = 1 or 0.5

We know,

x = nπ + (-1)ⁿθ

x = nπ + (-1)ⁿ\((\frac{π}2)\) or x = nπ + (-1)ⁿ\((\frac{π}6)\)

Therefore, the values of x are

\(\frac{π}6\),\(\frac{π}2\),\(\frac{5π}6\),π,\(\frac{13π}6\)

So, these values are obtained for different value of n from the equation

x = nπ + (-1)ⁿ\(\frac{π}2\) - \(\frac{π}6\),n ∈ Z.

So, Option B