Moderate -0.3 This is a straightforward application of the Pythagorean identity sin²x = 1 - cos²x to transform the equation, followed by solving a quadratic in cos x and finding angles from a calculator. Part (a) requires only direct substitution of a standard identity, and part (b) involves routine quadratic solving and inverse trig—both are standard C2 techniques with no novel problem-solving required.
7. (a) Show that
$$12 \sin ^ { 2 } x - \cos x - 11 = 0$$
may be expressed in the form
$$12 \cos ^ { 2 } x + \cos x - 1 = 0$$
(b) Hence, using trigonometry, find all the solutions in the interval \(0 \leqslant x \leqslant 360 ^ { \circ }\) of
$$12 \sin ^ { 2 } x - \cos x - 11 = 0$$
Give each solution, in degrees, to 1 decimal place.
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\(12\sin^2 x - \cos x - 11 = 0\) leading to \(12(1-\cos^2 x) - \cos x - 11 = 0\) and so \(12\cos^2 x + \cos x - 1 = 0\)
B1*
Replaces \(\sin^2 x\) by \((1-\cos^2 x)\) or replaces 11 by \(11(\sin^2 x + \cos^2 x)\) with no errors, giving printed answer including \(= 0\)
Part (b):
Answer
Marks
Guidance
Solve quadratic to obtain \(\cos x = \frac{1}{4}\) or \(-\frac{1}{3}\)
M1 A1
Both answers needed; allow 0.25 and awrt \(-0.33\)
\(x = 75.5, 109.5, 250.5, 284.5\)
M1 A1cao
Uses inverse cosine to obtain two correct values for \(x\) for their values of \(\cos x\); all four correct awrt; answers in radians are 1.3, 5.0, 1.9 and 4.4 — allow M1A0 for two or more correct answers
## Question 7:
### Part (a):
| $12\sin^2 x - \cos x - 11 = 0$ leading to $12(1-\cos^2 x) - \cos x - 11 = 0$ and so $12\cos^2 x + \cos x - 1 = 0$ | B1* | Replaces $\sin^2 x$ by $(1-\cos^2 x)$ or replaces 11 by $11(\sin^2 x + \cos^2 x)$ with no errors, giving printed answer including $= 0$ |
### Part (b):
| Solve quadratic to obtain $\cos x = \frac{1}{4}$ or $-\frac{1}{3}$ | M1 A1 | Both answers needed; allow 0.25 and awrt $-0.33$ |
| $x = 75.5, 109.5, 250.5, 284.5$ | M1 A1cao | Uses inverse cosine to obtain two correct values for $x$ for their values of $\cos x$; all four correct awrt; answers in radians are 1.3, 5.0, 1.9 and 4.4 — allow M1A0 for two or more correct answers |
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7. (a) Show that
$$12 \sin ^ { 2 } x - \cos x - 11 = 0$$
may be expressed in the form
$$12 \cos ^ { 2 } x + \cos x - 1 = 0$$
(b) Hence, using trigonometry, find all the solutions in the interval $0 \leqslant x \leqslant 360 ^ { \circ }$ of
$$12 \sin ^ { 2 } x - \cos x - 11 = 0$$
Give each solution, in degrees, to 1 decimal place.\\
\includegraphics[max width=\textwidth, alt={}, center]{e878227b-d625-4ef2-ac49-a9dc05c5321a-15_106_97_2615_1784}\\
\hfill \mbox{\textit{Edexcel C12 2014 Q7 [5]}}