Moderate -0.3 Part (a) is a straightforward algebraic manipulation using the identity cos²x = 1 - sin²x, requiring only substitution and rearrangement. Part (b) involves factorising a quadratic in sin x and solving basic trig equations in a given range. This is a standard C2 exercise with routine techniques and no problem-solving insight required, making it slightly easier than average.
2. (a) Show that the equation
$$5 \sin x = 1 + 2 \cos ^ { 2 } x$$
can be written in the form
$$2 \sin ^ { 2 } x + 5 \sin x - 3 = 0$$
(b) Solve, for \(0 \leqslant x < 360 ^ { \circ }\),
$$2 \sin ^ { 2 } x + 5 \sin x - 3 = 0$$
Correct method to change \(\cos^2 x\) into \(\sin^2 x\); must use \(\cos^2 x = 1 - \sin^2 x\)
\(2\sin^2 x + 5\sin x - 3 = 0\)
A1cso
Needs 3-term quadratic printed in any order with \(=0\) included
Part (b):
Answer
Marks
Guidance
Answer/Working
Marks
Guidance
\((2s-1)(s+3) = 0\) giving \(s =\)
M1
Attempt to solve given quadratic (usual rules for solving quadratics)
\([\sin x = -3\) has no solution\(]\) so \(\sin x = \frac{1}{2}\)
A1
Requires no incorrect work; for \(\sin x = \frac{1}{2}\) or \(x = \sin^{-1}\frac{1}{2}\)
\(\therefore x = 30°, 150°\)
B1, B1ft
B1 for 30°; 2nd B1 for \(180° - \alpha\); lose final B1 for answers in radians or extra solutions in range
## Question 2:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $5\sin x = 1 + 2(1 - \sin^2 x)$ | M1 | Correct method to change $\cos^2 x$ into $\sin^2 x$; must use $\cos^2 x = 1 - \sin^2 x$ |
| $2\sin^2 x + 5\sin x - 3 = 0$ | A1cso | Needs 3-term quadratic printed in any order with $=0$ included |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(2s-1)(s+3) = 0$ giving $s =$ | M1 | Attempt to solve given quadratic (usual rules for solving quadratics) |
| $[\sin x = -3$ has no solution$]$ so $\sin x = \frac{1}{2}$ | A1 | Requires no incorrect work; for $\sin x = \frac{1}{2}$ or $x = \sin^{-1}\frac{1}{2}$ |
| $\therefore x = 30°, 150°$ | B1, B1ft | B1 for 30°; 2nd B1 for $180° - \alpha$; lose final B1 for answers in radians or extra solutions in range |
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2. (a) Show that the equation
$$5 \sin x = 1 + 2 \cos ^ { 2 } x$$
can be written in the form
$$2 \sin ^ { 2 } x + 5 \sin x - 3 = 0$$
(b) Solve, for $0 \leqslant x < 360 ^ { \circ }$,
$$2 \sin ^ { 2 } x + 5 \sin x - 3 = 0$$
\hfill \mbox{\textit{Edexcel C2 2010 Q2 [6]}}