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8. (i) Solve, for $0 rown 0 < heta < rac{ heta}{ heta}$, the equation \[ \sin 3\theta - \sqrt{3} \cos 3\theta = 0 \] giving your answers in terms of $\pi$ - Edexcel - A-Level Maths Pure - Question 1 - 2014 - Paper 2
Question 1
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8. (i) Solve, for $0 rown 0 < heta < rac{ heta}{ heta}$, the equation \[ \sin 3\theta - \sqrt{3} \cos 3\theta = 0 \] giving your answers in terms of $\pi$. (... show full transcript
Worked Solution & Example Answer:8. (i) Solve, for $0 rown 0 < heta < rac{ heta}{ heta}$, the equation \[ \sin 3\theta - \sqrt{3} \cos 3\theta = 0 \] giving your answers in terms of $\pi$ - Edexcel - A-Level Maths Pure - Question 1 - 2014 - Paper 2
Step 1
Solve, for $0 < \theta < \pi$, the equation \$\sin 3\theta - \sqrt{3} \cos 3\theta = 0\$
Answer
To solve the equation, we can rewrite it as:
[ \sin 3\theta = \sqrt{3} \cos 3\theta ]
Dividing both sides by $\cos 3\theta$ (assuming $\cos 3\theta \neq 0$), we get:
[ \tan 3\theta = \sqrt{3} ]
The solutions for $\tan 3\theta = \sqrt{3}$ are given by:
[ 3\theta = \frac{\pi}{3} + n\pi, \quad n \in \mathbb{Z} ]
Thus,
[ \theta = \frac{\pi}{9} + \frac{n\pi}{3} ]
For $n=0$, $\theta = \frac{\pi}{9}$ is valid in the given range. For $n=1$, we get:
[ \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} $ is also valid.
Therefore, the solutions in terms of $\pi$ are: [ \frac{\pi}{9}, \frac{4\pi}{9}, \frac{7\pi}{9} ]
Step 2
Given that 4sin² x + cos x = 4 - k, 0 ≤ k ≤ 3 (a) find cos x in terms of k.
Answer
Starting with the equation:
[ 4\sin^{2} x + \cos x = 4 - k ]
We can express $\sin^{2} x$ using the Pythagorean identity $\sin^{2} x = 1 - \cos^{2} x$:
[ 4(1 - \cos^{2} x) + \cos x = 4 - k ]
Expanding gives:
[ 4 - 4\cos^{2} x + \cos x = 4 - k ]
Rearranging yields:
[ 4\cos^{2} x - \cos x + (k) = 0 ]
This is a quadratic equation in terms of $\cos x$. Using the quadratic formula, we have:
[ \cos x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{1 \pm \sqrt{1^2 - 4 * 4 * k}}{2 * 4} = \frac{1 \pm \sqrt{1 - 16k}}{8} ]
Step 3
When k = 3, find the values of x in the range 0 ≤ x < 360°
Answer
Substituting $k = 3$ into the equation we derived:
[ \cos x = \frac{1 \pm \sqrt{1 - 16 * 3}}{8} = \frac{1 \pm \sqrt{1 - 48}}{8} = \frac{1 \pm \sqrt{-47}}{8} ]
This indicates that there are no real solutions for $\cos x$ since the term inside the square root is negative. Therefore, there are no angles $x$ that satisfy the equation when $k = 3$ in the specified range.