A-Level Edexcel Maths Pure Applications of Differentiation: Solve for $0 \leq x

Solve for $0 \leq x < 360^\circ$, giving your answers in degrees to 1 decimal place, 3 sin$(x+45^\circ) = 2$ - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 2

Question 8

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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 2sin... 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$(x+45^\circ) = 2$ - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 2

Step 1

Solve for $0 \leq x < 360^\circ$, giving your answers in degrees to 1 decimal place, 3 sin$(x+45^\circ) = 2$

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Answer

To solve the equation, we first isolate sin $(x + 45^\circ)$ :

3 \sin(x + 45^\circ) = 2 \implies \sin(x + 45^\circ) = \frac{2}{3}

Next, we find the value of $(x + 45^\circ)$ using the inverse sine function:

(x + 45^\circ) = \sin^{-1}\left(\frac{2}{3}\right)

Calculating this gives us:

(x + 45^\circ) \approx 41.81^\circ

Since sine is positive in the first and second quadrants, we have another solution:

(x + 45^\circ) = 180^\circ - 41.81^\circ = 138.19^\circ

Now we isolate $x$ :

For the first solution: $$$ (not valid in the given range)$
For the second solution:

Thus, we need to adjust $x$ for periodicity: 3. Finding another angle in the allowed range: $$x = 356.81^\circ - 45^\circ = 311.81^\circ\

Finally, the valid solutions are:

\text{Answer: } 93.2^\circ, 311.8^\circ

Step 2

Find, for $0 \leq x < 2\pi$, all the solutions of 2sin$^2 x + 2 = 7\cos x$

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Answer

First, we rewrite the equation:

2\sin^2 x - 7\cos x + 2 = 0

Using the identity $\sin^2 x = 1 - \cos^2 x$ we convert the equation to:

2(1 - \cos^2 x) - 7\cos x + 2 = 0 \implies -2\cos^2 x - 7\cos x + 4 = 0

Multiplying through by -1 gives:

2\cos^2 x + 7\cos x - 4 = 0

Next, we apply the quadratic formula:

\cos x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 2 \cdot (-4)}}{2 \cdot 2}

Calculating the values:

\cos x = \frac{-7 \pm \sqrt{49 + 32}}{4} = \frac{-7 \pm \sqrt{81}}{4} = \frac{-7 \pm 9}{4}

Thus we find two potential solutions:

$\cos x = \frac{2}{4} = \frac{1}{2}$
$\cos x = \frac{-16}{4} = -4$ (not valid, since cosine values must be between -1 and 1)

From the first solution: $\cos x = \frac{1}{2}$ gives us:

Thus, the solutions in radians are:

$\text{Answer: } x \approx \frac{\pi}{3}, \frac{5\pi}{3}$

Solve for $0 \leq x < 360^\circ$, giving your answers in degrees to 1 decimal place, 3 sin$(x+45^\circ) = 2$ - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 2

Worked Solution & Example Answer:Solve for $0 \leq x < 360^\circ$, giving your answers in degrees to 1 decimal place, 3 sin$(x+45^\circ) = 2$ - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 2

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 2sin$^2 x + 2 = 7\cos x$

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