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 of $x$, in the interval $0 \leq x < 360^{\circ}$, for which $$9\cos^{2} x - 11\cos x + 3\sin^{2} x = 0$$ giving your answers to 1 decimal place. You must show clearly how you obtained your answers.

A-Level Edexcel Maths Pure Polynomials: Solve, for $0 \leq \theta

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 6

Question 2

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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 of $x$, in the ... show full transcript

Worked Solution & Example Answer: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 6

Step 1

sin(2θ - 30°) + 1 = 0.4

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Answer

Starting from the equation:

$\sin(2\theta - 30^{\circ}) + 1 = 0.4$

we can rearrange it to:

$\sin(2\theta - 30^{\circ}) = 0.4 - 1$ $\sin(2\theta - 30^{\circ}) = -0.6$

Now we find the angle for which the sine value is -0.6. Since sine is negative in the third and fourth quadrants, we can express this as:

The reference angle: $\theta_{ref} = \arcsin(0.6) \approx 36.9^{\circ}$
This gives us two angles for $2\theta - 30^{\circ}$ :

Third quadrant: $2\theta - 30^{\circ} = 180^{\circ} + 36.9^{\circ} \ 2\theta = 216.8^{\circ} \ \theta = 108.4^{\circ}$
Fourth quadrant: $2\theta - 30^{\circ} = 360^{\circ} - 36.9^{\circ} \ 2\theta = 323.1^{\circ} \ \theta = 161.55^{\circ}$

Thus, the solutions for $\theta$ in the interval $[0, 180^{\circ})$ are:

$\theta \approx 108.4^{\circ}$
$\theta \approx 161.6^{\circ}$ .

Step 2

Find all the values of x, in the interval 0 ≤ x < 360°, for which 9cos²x - 11cosx + 3sin²x = 0

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Answer

To solve the equation:

$9\cos^{2}x - 11\cos x + 3\sin^{2}x = 0$

we know that $\sin^{2}x = 1 - \cos^{2}x$ , which allows us to express everything in terms of $\cos x$ :

Substituting gives:

$9\cos^{2}x - 11\cos x + 3(1 - \cos^{2}x) = 0$ $9\cos^{2}x - 11\cos x + 3 - 3\cos^{2}x = 0$ $6\cos^{2}x - 11\cos x + 3 = 0$

Next, we can apply the quadratic formula to solve for $\cos x$ :

$\cos x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$ where $a = 6$ , $b = -11$ , and $c = 3$ :

$\cos x = \frac{11 \pm \sqrt{(-11)^{2} - 4 \cdot 6 \cdot 3}}{2 \cdot 6}$ $= \frac{11 \pm \sqrt{121 - 72}}{12}$ $= \frac{11 \pm \sqrt{49}}{12}$ $= \frac{11 \pm 7}{12}$

This gives us two values:

$\cos x = \frac{18}{12} = 1.5$ (not valid, as $\cos x$ must be in [ $-1$ , $1$ ])
$\cos x = \frac{4}{12} = \frac{1}{3}$

Now, to find $x$ , we have:

$x = \arccos\left(\frac{1}{3}\right) \approx 70.5^{\circ}$

The second solution in the range $[0, 360^{\circ})$ is given by:

$x = 360^{\circ} - 70.5^{\circ} \approx 289.5^{\circ}$

Thus, the solutions for $x$ are:

$x \approx 70.5^{\circ}$
$x \approx 289.5^{\circ}$ .

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 6

Worked Solution & Example Answer: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 6

sin(2θ - 30°) + 1 = 0.4

Find all the values of x, in the interval 0 ≤ x < 360°, for which 9cos²x - 11cosx + 3sin²x = 0

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