A-Level Edexcel Maths Pure Geometric Sequences & Series: Solve, for 0 ≤ x

Solve, for 0 ≤ x < 360°, (a) $\sin(x - 20^\circ) = -\frac{1}{\sqrt{2}}$ (b) $\cos 3x = -\frac{1}{2}$ - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 2

Question 2

$Solve,-for-0-≤-x-<-360°,---(a)--$\sin(x---20^\circ)-=--\frac{1}{\sqrt{2}}$---(b)--$\cos-3x-=--\frac{1}{2}$-Edexcel-A-Level Maths Pure-Question 2-2008-Paper 2.png$

Solve, for 0 ≤ x < 360°, (a) $\sin(x - 20^\circ) = -\frac{1}{\sqrt{2}}$ (b) $\cos 3x = -\frac{1}{2}$

Worked Solution & Example Answer:Solve, for 0 ≤ x < 360°, (a) $\sin(x - 20^\circ) = -\frac{1}{\sqrt{2}}$ (b) $\cos 3x = -\frac{1}{2}$ - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 2

Step 1

(a) $\sin(x - 20^\circ) = -\frac{1}{\sqrt{2}}$

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Answer

To solve the equation, we first recognize that the sine function is negative in the third and fourth quadrants.

Determine reference angle:
$\sin^{-1}\left(-\frac{1}{\sqrt{2}}\right) = -45^\circ$ ,
which corresponds to $45^\circ$ , as sine is periodic.
Find angles in the required quadrants:
- In the third quadrant:
  $x - 20^\circ = 180^\circ + 45^\circ = 225^\circ$
  $\Rightarrow x = 225^\circ + 20^\circ = 245^\circ$
- In the fourth quadrant:
  $x - 20^\circ = 360^\circ - 45^\circ = 315^\circ$
  $\Rightarrow x = 315^\circ + 20^\circ = 335^\circ$
The solutions in the interval $[0, 360^\circ)$ are:
- $x = 245^\circ, 335^\circ$ .

Step 2

(b) $\cos 3x = -\frac{1}{2}$

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Answer

To solve this equation, we first identify where the cosine function equals $-\frac{1}{2}$ .

Determine reference angles:
The angles where $\cos \theta = -\frac{1}{2}$ are $120^\circ$ and $240^\circ$ . So we have:
- $3x = 120^\circ + 360^\circ k$ ,
- $3x = 240^\circ + 360^\circ k$ ,
  for $k \in \mathbb{Z}$ .
Solve for $x$ :
- From $3x = 120^\circ$ , we get $x = 40^\circ + 120^\circ k$
- From $3x = 240^\circ$ , we get $x = 80^\circ + 120^\circ k$
Find values for $k$ such that $x$ is in the interval $[0, 120^\circ)$ (considering multiples of $120$ :
- For $k = 0$ :
  $x = 40^\circ$
  $x = 80^\circ$
- For $k = 1$ :
  $x = 160^\circ$
Collect valid solutions:
The solutions in the interval $[0, 360)$ are:
- $x = 40^\circ, 80^\circ, 160^\circ$ .

Solve, for 0 ≤ x < 360°, (a) $\sin(x - 20^\circ) = -\frac{1}{\sqrt{2}}$ (b) $\cos 3x = -\frac{1}{2}$ - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 2

Worked Solution & Example Answer:Solve, for 0 ≤ x < 360°, (a) $\sin(x - 20^\circ) = -\frac{1}{\sqrt{2}}$ (b) $\cos 3x = -\frac{1}{2}$ - Edexcel - A-Level Maths Pure - Question 2 - 2008 - Paper 2

(a) $\sin(x - 20^\circ) = -\frac{1}{\sqrt{2}}$

(b) $\cos 3x = -\frac{1}{2}$

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