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Given any expression of the form \(a\cos x + b\sin x\) it is better to rewrite it into one of the forms \(k\cos (x \pm \alpha )\) or \(k\sin (x \pm \alpha )\) before answering the question.

• its maximum value, which is given by \(k\) (since maximum value of sine/cosine is 1)
• its minimum value, which is given by \(- k\) (since minimum value of sine/cosine is -1)
• the roots of \(a\cos x + b\sin x = 0\) occur when either \(\sin (x \pm \alpha ) = 0\) or \(\cos (x \pm \alpha ) = 0\)

The information above can also be used to sketch the graph of \(y = a\cos x + b\sin x\).

\(\sqrt 5 \sin x + 2\cos x\) can be rewritten as \(3\cos (x - 0.841)\)

• 1. Write down the maximum and minimum values of \(\sqrt 5 \sin x + 2\cos x\) and the value(s) of \(x\) in the interval \(0 \le x \le 2\pi\) where these occur.
• 2. Determine the roots of the equation \(\sqrt 5 \sin x + 2\cos x = 0\) in the interval \(0 \le x \le 2\pi\)
• 3. Sketch the graph of \(y = \sqrt 5 \sin x + 2\cos x\) for \(0 \le x \le 2\pi\)
• 4. To find the maximum and minimum values, just have a look at the equation in its rewritten form. The maximum and minimum values are given by \(k\) and \(- k\) respectively.

So in this case, the maximum value is 3. This occurs when \(\cos (x - 0.841) = 1\) so \(x-0.841=\pi\) or \(2\pi\).

We'll ignore \(2\pi + 0.841\) as a possible solution because it's not in the domain specified. Which leaves us with \(x = 0.841\).

The minimum value is -3 which occurs when \(cos (x-0.841)= -1\) so \(x-0.841=\pi\) .

Solve this equation to find \(x = \pi + 0.841 = 3.983\)

• 5. The roots of the equation are where the graph crosses the \(x\)-axis. In this instance we can say that the roots occur when the cosine of the angle is zero.

\(\sqrt 5 \sin x + 2\cos x = 0 \Leftrightarrow 3\cos (x - 0.841) = 0\)

So \(\cos (x - 0.841) = 0\)

And \(x - 0.841 = \frac{\pi }{2}\) or \(\frac{{3\pi }}{2}\)

So \(x = \frac{\pi }{2} + 0.841\) or \(\frac{{3\pi }}{2} + 0.841\)

\(x = 2.412\,or\,5.553\)