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9. (i) Solve, for $0 \leq \theta < 180^{\circ}$, \[ \sin(2\theta - 30^{\circ}) + 1 = 0.4 \] giving your answers to 1 decimal place - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 5
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9. (i) Solve, for $0 \leq \theta < 180^{\circ}$, \[ \sin(2\theta - 30^{\circ}) + 1 = 0.4 \] giving your answers to 1 decimal place. (ii) Find all the values o... show full transcript
Worked Solution & Example Answer:9. (i) Solve, for $0 \leq \theta < 180^{\circ}$, \[ \sin(2\theta - 30^{\circ}) + 1 = 0.4 \] giving your answers to 1 decimal place - Edexcel - A-Level Maths Pure - Question 2 - 2013 - Paper 5
Step 1
Solve, for $0 \leq \theta < 180^{\circ}$, \[ \sin(2\theta - 30^{\circ}) + 1 = 0.4 \]
Answer
To solve the equation, begin by isolating the sine function:
[ \sin(2\theta - 30^{\circ}) = 0.4 - 1 = -0.6 ]
Next, we need to find the angle for which the sine value is -0.6. The reference angle is:
[ 2\theta - 30^{\circ} = \arcsin(-0.6) ] This gives us:
- Firstly, determine the value using a calculator: [ 2\theta - 30^{\circ} \approx -36.9^{\circ} ] (in the fourth quadrant)
- The second solution using the sine function in the second quadrant is: [ 2\theta - 30^{\circ} = 180^{\circ} + 36.9^{\circ} = 216.87^{\circ} ]
From here, we can solve for ( \theta ):
For the first solution: [ 2\theta = -36.9^{\circ} + 30^{\circ} = -6.9^{\circ} ] [ \theta = -3.45^{\circ} ] (not valid) [ 2\theta \approx 216.87^{\circ} + 30^{\circ} \approx 246.87^{\circ} ] Thus, [ \theta = \frac{246.87}{2} \approx 123.4^{\circ} ]
The valid solution in the range is: [ \theta \approx 123.4^{\circ} ]
Step 2
Find all the values of $x$, in the interval $0 \leq x < 360^{\circ}$, for which \[ 9\cos^2{x} - 11\cos{x} + 3\sin^2{x} = 0 \]
Answer
First, convert using the identity :
[ 9\cos^2{x} - 11\cos{x} + 3(1 - \cos^2{x}) = 0 ]
Simplifying this gives: [ (9 - 3)\cos^2{x} - 11\cos{x} + 3 = 0 ] [ 6\cos^2{x} - 11\cos{x} + 3 = 0 ]
Now, let , the equation becomes a quadratic: [ 6y^2 - 11y + 3 = 0 ]
Using the quadratic formula: [ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ] Where , , and : [ y = \frac{11 \pm \sqrt{(-11)^2 - 4 \cdot 6 \cdot 3}}{2 \cdot 6} ] [ = \frac{11 \pm \sqrt{121 - 72}}{12} ] [ = \frac{11 \pm \sqrt{49}}{12} ] [ = \frac{11 \pm 7}{12} ]
This gives us two solutions:
- [ y = \frac{18}{12} = 1.5 ] (not valid since )
- [ y = \frac{4}{12} = \frac{1}{3} ]
So we need to solve ( \cos{x} = \frac{1}{3} ):
The possible angles are:
- [ x = \arccos\left(\frac{1}{3}\right) \approx 70.5^{\circ} ]
- Second quadrant solution: [ x = 360^{\circ} - 70.5^{\circ} \approx 289.5^{\circ} ]
Therefore, the solutions for ( x ) are: [ x \approx 70.5^{\circ} ] and [ x \approx 289.5^{\circ} ]