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7. (i) Solve, for $0 \leq \theta < 360^\circ$, the equation $$9 \sin(\theta + 60^\circ) = 4$$ giving your answers to 1 decimal place - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 1

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7.-(i)-Solve,-for-$0-\leq-\theta-<-360^\circ$,-the-equation--$$9-\sin(\theta-+-60^\circ)-=-4$$-giving-your-answers-to-1-decimal-place-Edexcel-A-Level Maths Pure-Question 9-2014-Paper 1.png

7. (i) Solve, for $0 \leq \theta < 360^\circ$, the equation $$9 \sin(\theta + 60^\circ) = 4$$ giving your answers to 1 decimal place. You must show each step of you... show full transcript

Worked Solution & Example Answer:7. (i) Solve, for $0 \leq \theta < 360^\circ$, the equation $$9 \sin(\theta + 60^\circ) = 4$$ giving your answers to 1 decimal place - Edexcel - A-Level Maths Pure - Question 9 - 2014 - Paper 1

Step 1

Solve, for $0 \leq \theta < 360^\circ$, the equation $$9 \sin(\theta + 60^\circ) = 4$$

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Answer

To solve the equation, first isolate the sine term:

sin(θ+60)=49\sin(\theta + 60^\circ) = \frac{4}{9}

Using a calculator, we can find the angle:

θ+60=arcsin(49)\theta + 60^\circ = \arcsin\left(\frac{4}{9}\right)

Calculating:

arcsin(49)26.3877\arcsin\left(\frac{4}{9}\right) \approx 26.3877^\circ

We can also find the second angle in the range 0θ<3600 \leq \theta < 360^\circ:

θ+60=18026.3877=153.6123\theta + 60^\circ = 180^\circ - 26.3877^\circ = 153.6123^\circ

Now, let's isolate θ\theta:

  1. For the first angle:

θ=26.387760=33.6123\theta = 26.3877^\circ - 60^\circ = -33.6123^\circ (not in range)

  1. For the second angle:

θ=153.612360=93.6123\theta = 153.6123^\circ - 60^\circ = 93.6123^\circ

  1. Find the other equivalent angle:

θ+60=36026.3877\theta + 60^\circ = 360^\circ - 26.3877^\circ

So:

θ+60=326.6123\theta + 60^\circ = 326.6123^\circ

Thus,

θ=326.612360=266.6123\theta = 326.6123^\circ - 60^\circ = 266.6123^\circ

Finally, rounding to 1 decimal place, we have:

θ93.6,326.6\theta \approx 93.6^\circ, 326.6^\circ

Step 2

Solve, for $-\pi < x < \pi$, the equation $$2 \tan x - 3 \sin x = 0$$

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Answer

Rearranging the equation:

2tanx=3sinx2 \tan x = 3 \sin x

Using the identity tanx=sinxcosx\tan x = \frac{\sin x}{\cos x}, we have:

2sinxcosx=3sinx2 \frac{\sin x}{\cos x} = 3 \sin x

This simplifies to:

2sinx=3sinxcosx2 \sin x = 3 \sin x \cos x

Rearranging gives:

3sinxcosx2sinx=03 \sin x \cos x - 2 \sin x = 0

Factoring out sinx\sin x leads to:

sinx(3cosx2)=0\sin x(3 \cos x - 2) = 0

Setting each factor equal to zero provides:

  1. sinx=0\sin x = 0:

    • Solutions are: x=0,π,πx = 0, -\pi, \pi (but π-\pi and π\pi are outside the range)
  2. 3cosx2=03 \cos x - 2 = 0:

    • So: cosx=23\cos x = \frac{2}{3}
    • Taking the inverse cosine:
    • We find: x=arccos(23)0.8410x = \arccos\left(\frac{2}{3}\right) \approx 0.8410
    • The second solution in the given range: x=arccos(23)0.8410x = -\arccos\left(\frac{2}{3}\right) \approx -0.8410

Thus, the solutions to 2 decimal places are:

x0.84,0.84x \approx 0.84, -0.84

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