Solve, for $0 \leq \theta < \pi$, the equation $$\sin 3\theta - \sqrt{3} \cos 3\theta = 0$$ giving your answers in terms of $\pi$. (ii) Given that $$4\sin^{2} x + \cos x = 4 - k,\ 0 \leq k \leq 3$$ (a) find $\cos x$ in terms of $k$. (b) When $k = 3$, find the values of $x$ in the range $0 \leq x < 360^{\circ}$.

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Solve, for $0 \leq \theta < \pi$, the equation $$\sin 3\theta - \sqrt{3} \cos 3\theta = 0$$ giving your answers in terms of $\pi$ - Edexcel - A-Level Maths Pure - Question 2 - 2015 - Paper 2

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Solve, for $0 \leq \theta < \pi$, the equation $$\sin 3\theta - \sqrt{3} \cos 3\theta = 0$$ giving your answers in terms of $\pi$. (ii) Given that $$4\sin^{2} x +... show full transcript

Worked Solution & Example Answer:Solve, for $0 \leq \theta < \pi$, the equation $$\sin 3\theta - \sqrt{3} \cos 3\theta = 0$$ giving your answers in terms of $\pi$ - Edexcel - A-Level Maths Pure - Question 2 - 2015 - Paper 2

Step 1

Solve, for $0 \leq \theta < \pi$, the equation $$\sin 3\theta - \sqrt{3} \cos 3\theta = 0$$

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Answer

To solve the equation, we can rewrite it as:

$\frac{\sin 3\theta}{\cos 3\theta} = \sqrt{3}$

This means that:

$\tan 3\theta = \sqrt{3}$

The general solutions for $\tan \phi = \sqrt{3}$ occur at:

$\phi = \frac{\pi}{3} + n\pi, \quad n \in \mathbb{Z}$

Setting $\phi = 3\theta$ , we have:

$3\theta = \frac{\pi}{3} + n\pi$

To find $\theta$ , divide both sides by 3:

$\theta = \frac{\pi}{9} + \frac{n\pi}{3}$

We need $\theta$ to be in the interval $0 \leq \theta < \pi$ . For $n=0$ , we get:

$\theta = \frac{\pi}{9}$

For $n=1$ :

$\theta = \frac{\pi}{9} + \frac{\pi}{3} = \frac{\pi}{9} + \frac{3\pi}{9} = \frac{4\pi}{9}$

For $n=2$ :

$\theta = \frac{\pi}{9} + \frac{2\pi}{3} = \frac{\pi}{9} + \frac{6\pi}{9} = \frac{7\pi}{9}$

For $n=3$ , it exceeds $\pi$ . Therefore, the solutions in that interval are:

$\theta = \frac{\pi}{9}, \frac{4\pi}{9}, \frac{7\pi}{9}$

Step 2

Given that $$4\sin^{2} x + \cos x = 4 - k,\ 0 \leq k \leq 3$$ find $\cos x$ in terms of $k$.

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Answer

Rearranging the equation gives:

$4\sin^{2} x + \cos x + k - 4 = 0$

Using the identity $\sin^{2} x = 1 - \cos^{2} x$ , we substitute:

$4(1 - \cos^{2} x) + \cos x + k - 4 = 0$

This simplifies to:

$-4\cos^{2} x + \cos x + (k - 4) = 0$

Rearranging yields:

$4\cos^{2} x - \cos x - (k - 4) = 0$

Applying the quadratic formula:

$\cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 4$ , $b = -1$ , and $c = -(k - 4)$ gives:

$\cos x = \frac{1 \pm \sqrt{1 + 16(k - 4)}}{8}$

Thus, the final answer for $\cos x$ in terms of $k$ is:

$\cos x = \frac{1 \pm \sqrt{16k - 63}}{8}$

Step 3

When $k = 3$, find the values of $x$ in the range $0 \leq x < 360^{\circ}$

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Answer

Substituting $k = 3$ , we have:

$\cos x = \frac{1 \pm \sqrt{16(3) - 63}}{8} = \frac{1 \pm \sqrt{48 - 63}}{8} = \frac{1 \pm \sqrt{-15}}{8}$

Since this expression involves the square root of a negative number, $\cos x$ cannot have real values. Hence, there are no values of $x$ satisfying this equation in the range $0 \leq x < 360^{\circ}$ .

Solve, for $0 \leq \theta < \pi$, the equation $$\sin 3\theta - \sqrt{3} \cos 3\theta = 0$$ giving your answers in terms of $\pi$ - Edexcel - A-Level Maths Pure - Question 2 - 2015 - Paper 2

Worked Solution & Example Answer:Solve, for $0 \leq \theta < \pi$, the equation $$\sin 3\theta - \sqrt{3} \cos 3\theta = 0$$ giving your answers in terms of $\pi$ - Edexcel - A-Level Maths Pure - Question 2 - 2015 - Paper 2

Solve, for $0 \leq \theta < \pi$, the equation $$\sin 3\theta - \sqrt{3} \cos 3\theta = 0$$

Given that $$4\sin^{2} x + \cos x = 4 - k,\ 0 \leq k \leq 3$$ find $\cos x$ in terms of $k$.

When $k = 3$, find the values of $x$ in the range $0 \leq x < 360^{\circ}$

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