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Solve, for $0^{ ext{o}} < s < 180^{ ext{o}}$, the equation (a) $\sin (x + 10^{ ext{o}}) = \frac{\sqrt{3}}{2}$ - Edexcel - A-Level Maths Pure - Question 7 - 2005 - Paper 2

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Solve,-for-$0^{-ext{o}}-<-s-<-180^{-ext{o}}$,-the-equation--(a)-$\sin-(x-+-10^{-ext{o}})-=-\frac{\sqrt{3}}{2}$-Edexcel-A-Level Maths Pure-Question 7-2005-Paper 2.png

Solve, for $0^{ ext{o}} < s < 180^{ ext{o}}$, the equation (a) $\sin (x + 10^{ ext{o}}) = \frac{\sqrt{3}}{2}$. (b) $\cos 2x = 0.9$, giving your answers to 1 decima... show full transcript

Worked Solution & Example Answer:Solve, for $0^{ ext{o}} < s < 180^{ ext{o}}$, the equation (a) $\sin (x + 10^{ ext{o}}) = \frac{\sqrt{3}}{2}$ - Edexcel - A-Level Maths Pure - Question 7 - 2005 - Paper 2

Step 1

(a) $\sin (x + 10^{\text{o}}) = \frac{\sqrt{3}}{2}$

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Answer

To solve the equation, we first find the reference angle for which the sine value equals 32\frac{\sqrt{3}}{2}. This angle is 60o60^{\text{o}}.

  1. Since sin⁡θ=sin⁡(180o−θ)\sin \theta = \sin (180^{\text{o}} - \theta), we have two potential angles:

    • x+10o=60ox + 10^{\text{o}} = 60^{\text{o}}
    • x+10o=120ox + 10^{\text{o}} = 120^{\text{o}}

  2. Rearranging these gives:

    • For the first equation:
      x=60o−10o=50ox = 60^{\text{o}} - 10^{\text{o}} = 50^{\text{o}}
    • For the second equation:
      x=120o−10o=110ox = 120^{\text{o}} - 10^{\text{o}} = 110^{\text{o}}

Thus, the solutions for part (a) are:

  • x=50ox = 50^{\text{o}}
  • x=110ox = 110^{\text{o}}

Step 2

(b) $\cos 2x = 0.9$

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Answer

To tackle this equation, we will first rewrite it:

  1. We will subtract 0.9 from both sides: cos⁡2x−0.9=0\cos 2x - 0.9 = 0

  2. The next step involves finding the reference angle:

    • 2x=cos⁡−1(0.9)2x = \cos^{-1}(0.9).

    Computing this gives: 2x=25.84o2x = 25.84^{\text{o}}

  3. The general solution for the cosine function is:

    • 2x=360o−25.84o2x = 360^{\text{o}} - 25.84^{\text{o}} (to account for the positive angle)
    • Thus, we also have:
    • 2x=25.84o2x = 25.84^{\text{o}} and 2x=334.16o2x = 334.16^{\text{o}}

  4. Now, we divide both sides by 2:

    • From 2x=25.84o2x = 25.84^{\text{o}}:
      x=25.84o2=12.92ox = \frac{25.84^{\text{o}}}{2} = 12.92^{\text{o}}
    • From 2x=334.16o2x = 334.16^{\text{o}}:
      x=334.16o2=167.08ox = \frac{334.16^{\text{o}}}{2} = 167.08^{\text{o}}

Thus, the two solutions for part (b) are:

  • x=12.9ox = 12.9^{\text{o}} (to 1 decimal place)
  • x=167.1ox = 167.1^{\text{o}} (to 1 decimal place)

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