Photo AI

7. (i) Solve, for $0 \, \leq \, x \, < \, \frac{\pi}{2}$, the equation $$4 \sin x = \sec x$$ (ii) Solve, for $0 \leq \theta < 360^\circ$, the equation $$5 \sin \theta - 5 \cos \theta = 2$$ giving your answers to one decimal place - Edexcel - A-Level Maths Pure - Question 9 - 2018 - Paper 2

Question icon

Question 9

7.-(i)-Solve,-for-$0-\,-\leq-\,-x-\,-<-\,-\frac{\pi}{2}$,-the-equation--$$4-\sin-x-=-\sec-x$$--(ii)-Solve,-for-$0-\leq-\theta-<-360^\circ$,-the-equation--$$5-\sin-\theta---5-\cos-\theta-=-2$$--giving-your-answers-to-one-decimal-place-Edexcel-A-Level Maths Pure-Question 9-2018-Paper 2.png

7. (i) Solve, for $0 \, \leq \, x \, < \, \frac{\pi}{2}$, the equation $$4 \sin x = \sec x$$ (ii) Solve, for $0 \leq \theta < 360^\circ$, the equation $$5 \sin \t... show full transcript

Worked Solution & Example Answer:7. (i) Solve, for $0 \, \leq \, x \, < \, \frac{\pi}{2}$, the equation $$4 \sin x = \sec x$$ (ii) Solve, for $0 \leq \theta < 360^\circ$, the equation $$5 \sin \theta - 5 \cos \theta = 2$$ giving your answers to one decimal place - Edexcel - A-Level Maths Pure - Question 9 - 2018 - Paper 2

Step 1

Solve, for $0 \leq x < \frac{\pi}{2}$, the equation $4 \sin x = \sec x$

96%

114 rated

Answer

To solve the equation, we start by rewriting it as: 4sin⁡x=1cos⁡x4 \sin x = \frac{1}{\cos x}

Multiplying both sides by (\cos x) gives: 4sin⁡xcos⁡x=14 \sin x \cos x = 1

Using the double angle identity (\sin(2x) = 2 \sin x \cos x), we can rewrite: 2sin⁡(2x)=12 \sin(2x) = 1

Thus, sin⁡(2x)=12\sin(2x) = \frac{1}{2}

The general solution for (\sin(2x) = \frac{1}{2}) is: 2x=π6+2kπor2x=5π6+2kπ(k∈Z)2x = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad 2x = \frac{5\pi}{6} + 2k\pi\quad (k \in \mathbb{Z})

Solving these, we find:

  1. (x = \frac{\pi}{12} + k\pi)
  2. (x = \frac{5\pi}{12} + k\pi)

Since we are restricted to 0≤x<π20 \leq x < \frac{\pi}{2}, the only solutions in this range are:

  • (x = \frac{\pi}{12})
  • (x = \frac{5\pi}{12})

Step 2

Solve, for $0 \leq \theta < 360^\circ$, the equation $5 \sin \theta - 5 \cos \theta = 2$

99%

104 rated

Answer

We start by rewriting the equation: 5sin⁡θ−5cos⁡θ=25 \sin \theta - 5 \cos \theta = 2

Dividing both sides by 5, we obtain: sin⁡θ−cos⁡θ=25\sin \theta - \cos \theta = \frac{2}{5}

We can rearrange this to find: sin⁡θ=cos⁡θ+25\sin \theta = \cos \theta + \frac{2}{5}

Now, squaring both sides leads to: (sin⁡θ)2=(cos⁡θ+25)2(\sin \theta)^2 = (\cos \theta + \frac{2}{5})^2

Substituting (\sin^2 \theta + \cos^2 \theta = 1) gives: 1−cos⁡2θ=(cos⁡θ+25)21 - \cos^2 \theta = (\cos \theta + \frac{2}{5})^2

Expanding and rearranging results in: 0=2cos⁡2θ+45cos⁡θ+(425−1)0 = 2\cos^2 \theta + \frac{4}{5}\cos \theta + \left(\frac{4}{25}-1\right)

This standard quadratic can be solved using the quadratic formula: cos⁡θ=−b±b2−4ac2a\cos \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} with (a = 2, b = \frac{4}{5}, c = \left(\frac{4}{25}-1\right)).

Finding solutions using the quadratic formula will yield two values for (\theta) in the interval (0 \leq \theta < 360^\circ) giving:

  • Approximate values of (\theta \approx 61.4^\circ) and (\theta \approx 208.6^\circ).

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;