6. f(x) = 12 cos x - 4 sin x - Edexcel - A-Level Maths Pure - Question 8 - 2006 - Paper 5

Using the expression we found for f(x):

$12 cos x - 4 sin x = R cos(x + α)$

Setting:

$R = 4\sqrt{10}$ and $α = 18.43°$

Thus we reformulate the equation as:

$R cos(x + α) = 7$ $4\sqrt{10} cos(x + 18.43°) = 7$

To isolate cos, we divide by 4√10:

$cos(x + 18.43°) = \frac{7}{4\sqrt{10}} \approx 0.5534$

To find x, we calculate:

$x + 18.43° = cos^{-1}(0.5534) \approx 56.4°$

Next, we find the principal value:

$x = 56.4° - 18.43° \approx 37.97°$

And since cosine is positive in the fourth quadrant, the other solution is:

$x + 18.43° = 360° - 56.4° = 303.6°$ $x = 303.6° - 18.43° \approx 285.17°$

Thus the answers are approximately:

• x = 38.0°
• x = 285.2°.

The minimum value of the expression $12 cos x - 4 sin x$ is given by:

$-\sqrt{R^2} = -\sqrt{160} \approx -12.649$

So the minimum value is approximately -12.65.

The minimum value occurs when:

$x + 18.43° = 180° + 18.43° = 198.43°$

Thus, $x = 198.43° - 18.43° = 180°$.

This corresponds to the minimum value hence the smallest positive x occurring at:

$x \approx 161.57°$

Therefore, to two decimal places, the answer is:

• x = 161.57°.