Moderate -0.3 This is a standard C2 trigonometric equation requiring the Pythagorean identity (cos²x = 1 - sin²x) to convert to a quadratic in sin x, followed by routine algebraic manipulation to reach the given form and solve. The 'show that' part guides students through the algebra, making it slightly easier than average but still requiring multiple steps and careful execution.
8. (a) Show that the equation
$$\cos ^ { 2 } x = 8 \sin ^ { 2 } x - 6 \sin x$$
can be written in the form
$$( 3 \sin x - 1 ) ^ { 2 } = 2$$
(b) Hence solve, for \(0 \leqslant x < 360 ^ { \circ }\),
$$\cos ^ { 2 } x = 8 \sin ^ { 2 } x - 6 \sin x$$
giving your answers to 2 decimal places.
E.g. \(9\sin^{2}x-6\sin x=1\) or \(9\sin^{2}x-6\sin x-1=0\) or \(9\sin^{2}x-6\sin x+1=2\)
M1
So \(9\sin^{2}x-6\sin x+1=2\) or \((3\sin x-1)^{2}-2=0\) so \((3\sin x-1)^{2}=2\) or \(2=(3\sin x-1)^{2}\) *
A1cso*
Way 2: Expands \((3\sin x-1)^{2}=2\) and uses quadratic formula on 3TQ
B1, M1
(b) Way 1: \((3\sin x-1)=(\pm)\sqrt{2}\)
M1
\(\sin x=\frac{1\pm\sqrt{2}}{3}\) or awrt 0.8047 and awrt -0.1381
A1
\(x=53.58, 126.42\) (or 126.41), 352.06, 187.94
dM1 A1
Notes:
- (a) Way 1:
- B1: Uses \(\cos^{2}x=1-\sin^{2}x\)
- M1: Collects \(\sin^{2}x\) terms to form a three term quadratic or into a suitable completed square format. May be sign slips in the collection of terms.
- A1*: cso This needs an intermediate step from 3 term quadratic and no errors in answer and printed answer stated but allow \(2=(3\sin x-1)^{2}\). If sin is used throughout instead of sinx it is A0.
- Way 2:
- B1: Needs correct expansion and split
- M1: Collects \(1-\sin^{2}x\) together
- A1*: Conclusion and no errors seen
- (b) M1: Square roots both sides(Way 1), or expands and uses quadratic formula (Way 2) Attempts at factorization after expanding are M0; A1: Both correct answers for sinx (need plus and minus). Need not be simplified; dM1: Uses inverse sin to give one of the given correct answers; 1st A1: Need two correct angles (allow awrt) Note that the scheme allows 126.41 in place of 126.42 though 126.42 is preferred; A1: All four solutions correct (Extra solutions in range lose this A mark, but outside range - ignore); (Premature approximation – in the final three marks lose first A1 then ft other angles for second A mark); Do not require degrees symbol for the marks
- Special case: Working in radians
- M1A1A0 for the correct 0.94, 2.21, 6.14, 3.28
| (a) **Way 1:** $1-\sin^{2}x=8\sin^{2}x-6\sin x$ | B1 |
|---|---|
| E.g. $9\sin^{2}x-6\sin x=1$ or $9\sin^{2}x-6\sin x-1=0$ or $9\sin^{2}x-6\sin x+1=2$ | M1 |
| So $9\sin^{2}x-6\sin x+1=2$ or $(3\sin x-1)^{2}-2=0$ so $(3\sin x-1)^{2}=2$ or $2=(3\sin x-1)^{2}$ * | A1cso* |
| **Way 2:** Expands $(3\sin x-1)^{2}=2$ and uses quadratic formula on 3TQ | B1, M1 |
| (b) **Way 1:** $(3\sin x-1)=(\pm)\sqrt{2}$ | M1 |
| $\sin x=\frac{1\pm\sqrt{2}}{3}$ or awrt 0.8047 and awrt -0.1381 | A1 |
| $x=53.58, 126.42$ (or 126.41), 352.06, 187.94 | dM1 A1 |
**Notes:**
- **(a) Way 1:**
- B1: Uses $\cos^{2}x=1-\sin^{2}x$
- M1: Collects $\sin^{2}x$ terms to form a three term quadratic or into a suitable completed square format. May be sign slips in the collection of terms.
- A1*: cso This needs an intermediate step from 3 term quadratic and no errors in answer and printed answer stated but allow $2=(3\sin x-1)^{2}$. If sin is used throughout instead of sinx it is A0.
- **Way 2:**
- B1: Needs correct expansion and split
- M1: Collects $1-\sin^{2}x$ together
- A1*: Conclusion and no errors seen
- **(b)** M1: Square roots both sides(Way 1), or expands and uses quadratic formula (Way 2) Attempts at factorization after expanding are M0; **A1:** Both correct answers for sinx (need plus and minus). Need not be simplified; **dM1:** Uses inverse sin to give one of the given correct answers; **1st A1:** Need two correct angles (allow awrt) Note that the scheme allows 126.41 in place of 126.42 though 126.42 is preferred; **A1:** All four solutions correct (Extra solutions in range lose this A mark, but outside range - ignore); **(Premature approximation –** in the final three marks lose first A1 then ft other angles for second A mark)**; Do not require degrees symbol for the marks**
- **Special case: Working in radians**
- M1A1A0 for the correct 0.94, 2.21, 6.14, 3.28
8. (a) Show that the equation
$$\cos ^ { 2 } x = 8 \sin ^ { 2 } x - 6 \sin x$$
can be written in the form
$$( 3 \sin x - 1 ) ^ { 2 } = 2$$
(b) Hence solve, for $0 \leqslant x < 360 ^ { \circ }$,
$$\cos ^ { 2 } x = 8 \sin ^ { 2 } x - 6 \sin x$$
giving your answers to 2 decimal places.
\hfill \mbox{\textit{Edexcel C2 2017 Q8 [8]}}