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Solve for $0 \leq x < 360^\circ$, giving your answers in degrees to 1 decimal place, \[ 3 \sin(x + 45^\circ) = 2 \] Find, for $0 \leq x < 2\pi$, all the solutions of \[ 2 \sin^2 x + 2 = 7 \cos x \] giving your answers in radians - Edexcel - A-Level Maths Pure - Question 9 - 2011 - Paper 2
Question 9
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Solve for $0 \leq x < 360^\circ$, giving your answers in degrees to 1 decimal place, \[ 3 \sin(x + 45^\circ) = 2 \] Find, for $0 \leq x < 2\pi$, all the solutions o... show full transcript
Worked Solution & Example Answer:Solve for $0 \leq x < 360^\circ$, giving your answers in degrees to 1 decimal place, \[ 3 \sin(x + 45^\circ) = 2 \] Find, for $0 \leq x < 2\pi$, all the solutions of \[ 2 \sin^2 x + 2 = 7 \cos x \] giving your answers in radians - Edexcel - A-Level Maths Pure - Question 9 - 2011 - Paper 2
Step 1
Solve for $0 \leq x < 360^\circ$, giving your answers in degrees to 1 decimal place
Answer
To solve the equation:
[ 3 \sin(x + 45^\circ) = 2 ]
First, divide both sides by 3:
[ \sin(x + 45^\circ) = \frac{2}{3} ]
To find the angle, take the inverse sine:
[ x + 45^\circ = \sin^{-1}\left(\frac{2}{3}\right) ]
Calculating ( \sin^{-1}\left(\frac{2}{3}\right) ) gives approximately 41.8 degrees:
[ x + 45^\circ = 41.8103^\circ ]
Thus, solving for ( x ):
[ x = 41.8103^\circ - 45^\circ = -3.1897^\circ ]
Since this is negative, we add 360 degrees:
[ x \approx 356.8103^\circ ]
The sine function is periodic, so also consider the reference angle:
[ x + 45^\circ = 180^\circ - 41.8103^\circ ]
Calculating gives:
[ x + 45^\circ \approx 138.1897^\circ \Rightarrow x \approx 93.1897^\circ ]
Finally, the solutions are:
- (to 1 decimal place: 93.2 degrees)
- (to 1 decimal place: 356.8 degrees)
Step 2
Find, for $0 \leq x < 2\pi$, all the solutions of $2 \sin^2 x + 2 = 7 \cos x$
Answer
Start by rearranging the equation:
[ 2 \sin^2 x + 2 - 7 \cos x = 0 ]
Substituting ( \sin^2 x = 1 - \cos^2 x ) leads to:
[ 2(1 - \cos^2 x) + 2 - 7 \cos x = 0 ]
This simplifies to:
[ -2 \cos^2 x - 7 \cos x + 4 = 0 ]
Mulitplying by -1:
[ 2 \cos^2 x + 7 \cos x - 4 = 0 ]
Using the quadratic formula, ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ):
Where ( a = 2, b = 7, c = -4 ):
[ \cos x = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 2 \cdot (-4)}}{2 \cdot 2} ] [ \cos x = \frac{-7 \pm \sqrt{49 + 32}}{4} ] [ \cos x = \frac{-7 \pm \sqrt{81}}{4} ] [ \cos x = \frac{-7 \pm 9}{4} ]
This gives us two cases:
- [ \cos x = \frac{2}{4} = \frac{1}{2} ]
- [ \cos x = \frac{-16}{4} = -4 \text{ (not possible since } \cos x\text{ must be in } [-1, 1]) ]
From ( \cos x = \frac{1}{2} ), we find the angles:
[ x = \frac{\pi}{3},; x = \frac{5\pi}{3} ]
Thus, the solutions are: