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Solve for $0 \leq x < 360^\circ$, giving your answers in degrees to 1 decimal place, 3 sin$(x + 45^\circ) = 2$ - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 2

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Solve-for-$0-\leq-x-<-360^\circ$,-giving-your-answers-in-degrees-to-1-decimal-place,-3-sin$(x-+-45^\circ)-=-2$-Edexcel-A-Level Maths Pure-Question 8-2011-Paper 2.png

Solve for $0 \leq x < 360^\circ$, giving your answers in degrees to 1 decimal place, 3 sin$(x + 45^\circ) = 2$. Find, for $0 \leq x < 2\pi$, all the solutions of 2s... show full transcript

Worked Solution & Example Answer:Solve for $0 \leq x < 360^\circ$, giving your answers in degrees to 1 decimal place, 3 sin$(x + 45^\circ) = 2$ - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 2

Step 1

Solve for $0 \leq x < 360^\circ$, giving your answers in degrees to 1 decimal place, 3 sin$(x + 45^\circ) = 2$

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Answer

To solve the equation, start by isolating the sine term:

3sin⁡(x+45∘)=2⇒sin⁡(x+45∘)=233 \sin(x + 45^\circ) = 2 \Rightarrow \sin(x + 45^\circ) = \frac{2}{3}

Next, we apply the inverse sine function:

x+45∘=sin⁡−1(23) x + 45^\circ = \sin^{-1}\left(\frac{2}{3}\right)

Calculating this value,

sin⁡−1(23)≈41.8∘\sin^{-1}\left(\frac{2}{3}\right) \approx 41.8^\circ

Thus, we have:

  1. First solution:
x+45∘=41.8∘  ⟹  x=41.8∘−45∘≈−3.2∘ (nothinspacevalid)x + 45^\circ = 41.8^\circ \implies x = 41.8^\circ - 45^\circ \approx -3.2^\circ\ (not hinspace valid)
  1. Second solution:
x+45∘=180∘−41.8∘  ⟹  x=138.2∘x + 45^\circ = 180^\circ - 41.8^\circ \implies x = 138.2^\circ
  1. Third solution:
x+45∘=360∘+41.8∘  ⟹  x=360∘−45∘+41.8∘=356.8∘x + 45^\circ = 360^\circ + 41.8^\circ \implies x = 360^\circ - 45^\circ + 41.8^\circ = 356.8^\circ

In summary, the valid solutions for 0≤x<360∘0 \leq x < 360^\circ are:

{138.2∘,356.8∘}\{138.2^\circ, 356.8^\circ\}

Step 2

Find, for $0 \leq x < 2\pi$, all the solutions of 2sin$^2 x + 2 = 7$cos$x$

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Answer

Rearranging the equation gives us:

2sin⁡2x−7cos⁡x+2=02\sin^2 x - 7\cos x + 2 = 0

To solve this, we can use the identity sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x:

2(1−cos⁡2x)−7cos⁡x+2=02(1 - \cos^2 x) - 7\cos x + 2 = 0

Upon simplifying, we get:

−2cos⁡2x−7cos⁡x+4=0-2\cos^2 x - 7\cos x + 4 = 0

Multiplying through by -1 yields:

2cos⁡2x+7cos⁡x−4=02\cos^2 x + 7\cos x - 4 = 0

Using the quadratic formula, where a=2a = 2, b=7b = 7, and c=−4c = -4:

cos⁡x=−b±b2−4ac2a\cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Calculating:

cos⁡x=−7±49+324=−7±94\cos x = \frac{-7 \pm \sqrt{49 + 32}}{4} = \frac{-7 \pm 9}{4}

This results in:

  1. First root: cos⁡x=24=12\cos x = \frac{2}{4} = \frac{1}{2}

  2. Second root:

For the valid solution, cos⁡x=12\cos x = \frac{1}{2} gives:

  • x=π3x = \frac{\pi}{3} (1st quadrant)
  • x=5π3x = \frac{5\pi}{3} (4th quadrant)

Thus, the solutions for 0≤x<2π0 \leq x < 2\pi are:

{π3,5π3}\left\{ \frac{\pi}{3}, \frac{5\pi}{3} \right\}

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