Let me see if I can help clarify *how *to get to the correct answer.

First, draw a circle of radius R on a piece of paper with x & y axes through the center. Pick some arbitrary angle in the 1st quadrant and label it Θ. Next using this angle construct a right triangle from the x-axis, such that x = A and y = B.

Your problem asks to find the maximum value of √3 sin(x) + cos(x) = ? maximum

Let's put this equation in the general form of A sin(x) + B cos(x) = ?

Multiply top and bottom by R so that R/R [A sin(x) + B cos(x)] = ?

We get away with this because R/R = 1.

Now bring the R in the denominator inside and write: R*[A/R sin(x) + B/R cos(x)] = ?

From the picture that we drew with the circle and the angle we can see that:

sin(θ) = B/R and cos(Θ) = A/R, where R ^{2 }= A ^{2 }+ B ^{2}

Substituting into the equation above we can now write R*[cos(θ)*sin(x) + sin(Θ)cos(x)] = R*[sin(x + Θ)]

This takes on a maximum value when sin(x+ Θ)] = 1,

or when x + Θ = pi/2. then x = pi/2 - Θ.

Θ = sin ^{-1} (B/R)

For your problem A = √3, B = 1, and R = √(A ^{2} + B ^{2) }

so that R = 2 and B/R = ½

sin(Θ) = 1/2 when Θ = pi/6

Finally, x = pi/2 - pi/6 = pi/3. This is your maximum value for x.

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