Question

7. The value of x for the minimum
value of √3 cos x +sin x is​

Answer 1

Question :- The value of x for the minimum value of (√3 cos x +sin x) is ?

Solution :-

→ (√3 cos x +sin x)

Multiply and divide the equation by 2, we get,

→ (2/2)(√3 cos x +sin x)

Taking (1/2) inside now,

→ 2(√3/2 * cos x + 1/2 * sin x)

Putting :-

• √3/2 = cos 30°
• √3/2 = cos 30° 1/2 = sin 30°

→ 2(cos x * cos 30° + sin x * sin 30°.)

Now using :-

• cosA * cosB + sinA * sinB = cos(A - B)

→ 2 * cos( x - 30°)

Now, we know that,

• Minimum value of cos θ is (-1) when θ = 180 ˚.

So, 2 * cos( x - 30°) will be minimum when cos(x - 30°) is equal to (-1).

Therefore,

→ cos(x - 30°) = cos180°

→ x - 30° = 180°

→ x = 210°. (Ans.)

Hence, the value of x will be 210°.

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

$\sf{Let \: \: f(x) = \sqrt{3} \cos x + \sin x \: \: }$

Differentiating both sides with respect to x two times

$\sf{ {f} \: {'}(x) = - \sqrt{3} \sin x + \cos x \: \: }$

$\sf{ {f} \: {''}(x) = - ( \sqrt{3} \cos x + \sin x \:) \: }$

For extremum value of f(x) we have

$\sf{ {f} \: {'}(x) = 0\: }$

$\implies \: \sf{ - \sqrt{3} \sin x + \cos x \: = 0 \: }$

$\implies \: \sf{ \sqrt{3} \sin x = \cos x \: \: }$

$\implies \: \sf{ \cot x \: = \sqrt{3} \: }$

$\implies \: \sf{ x = \: {30}^{ \circ} , {210}^{ \circ}\: }$

Now

$\sf{ {f} \: {''}( {30}^{ \circ} ) = - ( \sqrt{3} \cos {30}^{ \circ}+ \sin {30}^{ \circ} \:) = - 2 < 0 \: }$

So f(x) has a maximum value at x = 30°

Again

$\sf{ {f} \: {''}( {210}^{ \circ} ) = - ( \sqrt{3} \cos {210}^{ \circ}+ \sin {210}^{ \circ} \:) = 2 > 0 \: }$

So f(x) has a minimum value at x = 210°

The minimum value is

$\sf{ {f}( {210}^{ \circ} ) = ( \sqrt{3} \cos {210}^{ \circ}+ \sin {210}^{ \circ} \:) = - 2 \: }$

RESULT

For √3 cos x +sin x the minimum value occurs at x = 210° and the minimum value is - 2