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Solve, for $0 \, \leq \, s \, < \, 180^\circ$, the equation (a) $\sin(x + 10^\circ) = \frac{\sqrt{3}}{2}$ - Edexcel - A-Level Maths Pure - Question 7 - 2005 - Paper 2
Question 7
Solve, for $0 \, \leq \, s \, < \, 180^\circ$, the equation (a) $\sin(x + 10^\circ) = \frac{\sqrt{3}}{2}$. (b) $\cos 2x = 0.9$, giving your answers to 1 decimal pl... show full transcript
Worked Solution & Example Answer:Solve, for $0 \, \leq \, s \, < \, 180^\circ$, the equation (a) $\sin(x + 10^\circ) = \frac{\sqrt{3}}{2}$ - Edexcel - A-Level Maths Pure - Question 7 - 2005 - Paper 2
Step 1
(a) $\sin(x + 10^\circ) = \frac{\sqrt{3}}{2}$
Answer
To solve the equation, we first identify the values of the angle whose sine is . This occurs at angles:
- (since sine is positive in the first and second quadrants).
Next, we set up two equations based on these angles:
-
[ x + 10^\circ = 60^\circ ] [ x = 60^\circ - 10^\circ = 50^\circ ]
-
[ x + 10^\circ = 120^\circ ] [ x = 120^\circ - 10^\circ = 110^\circ ]
Thus, the solutions for part (a) are:
Step 2
(b) $\cos 2x = 0.9$
Answer
To solve for , we first express the equation in terms of a simpler variable. We use the fact that implies:
-
[ 2x = \cos^{-1}(0.9) ]
Using a calculator, we find: [ 2x \approx 25.8^\circ ] Therefore:\n [ x \approx \frac{25.8^\circ}{2} \approx 12.9^\circ ] -
The cosine function is also positive in the fourth quadrant. Thus, the solution is: [ 2x = 360^\circ - 25.8^\circ = 334.2^\circ ] Hence:\n [ x \approx \frac{334.2^\circ}{2} \approx 167.1^\circ ]
Since only the domain of is considered, we select:
- (but this is outside the required range, so ignore it)
Therefore, the final answer for part (b) is: