Free · no sign-up to startฟรี ไม่ต้องสมัครก็เริ่มได้
A-Level Pure 2 · Edexcel IAL WMA12 & CAIE 9709 · Topic pack

Topic pack · Trigonometry แพ็กฝึกเฉพาะหัวข้อ · เอกลักษณ์และสมการตรีโกณ

10 questions 10 ข้อ 55 marks 55 คะแนน 70 min 70 นาที Identities & equations, easy → hard

Ten questions on one topic, ordered easy → hard. The last three are full exam difficulty. Open the mark schemes only after a real attempt. สิบข้อหัวข้อเดียว เรียงง่าย → ยาก สามข้อสุดท้ายคือระดับข้อสอบจริง เปิดมาร์คสกีมหลังลองทำจริงเท่านั้น

⏱ 70:00 Scoreคะแนนรวม: –
Q1. 3 marks

Solve, for $0^\circ \le x < 360^\circ$, the equation $\tan x = \sqrt{3}$.

Mark schemeshow ▾
M1 principal value $\tan^{-1}\sqrt{3}=60^\circ$  A1 $x=60^\circ$  A1 $x=240^\circ$ (adding $180^\circ$; no others in range).
Q2. 4 marks

Solve, for $0^\circ \le x < 360^\circ$, the equation $3\sin x = \sqrt{3}\cos x$.

Mark schemeshow ▾
M1 divide both sides by $\cos x$ and use $\tan x=\dfrac{\sin x}{\cos x}$  A1 $\tan x=\dfrac{\sqrt{3}}{3}$  A1 $x=30^\circ$  A1 $x=210^\circ$.
Q3. 4 marks

Solve, for $0^\circ \le x < 360^\circ$, the equation $\cos(x-30^\circ) = \dfrac{1}{2}$.

Mark schemeshow ▾
M1 let $u=x-30^\circ$, widen the interval: $-30^\circ \le u < 330^\circ$  A1 $u=60^\circ$ or $u=300^\circ$ ($u=-60^\circ$ rejected, out of range)  A1 $x=90^\circ$  A1 $x=330^\circ$.
Q4. 5 marks

Solve, for $0^\circ \le x < 360^\circ$, the equation $2\sin^2 x + \sin x - 1 = 0$.

Mark schemeshow ▾
M1 factorise: $(2\sin x-1)(\sin x+1)=0$  A1 $\sin x=\dfrac12$ or $\sin x=-1$  A1 $x=30^\circ$  A1 $x=150^\circ$  A1 $x=270^\circ$.
Q5. 5 marks

Solve, for $0^\circ \le x < 360^\circ$, the equation $\sin 2x = \dfrac{\sqrt{3}}{2}$.

Mark schemeshow ▾
M1 widen the interval: $0^\circ \le 2x < 720^\circ$  A1 $2x=60^\circ,\ 120^\circ$  M1 add $360^\circ$: $2x=420^\circ,\ 480^\circ$  A1 $x=30^\circ,\ 60^\circ$  A1 $x=210^\circ,\ 240^\circ$.
Q6. 6 marks

(a) Show that the equation $2\cos^2 x + 3\sin x = 3$ can be written in the form $2\sin^2 x - 3\sin x + 1 = 0$. (b) Hence solve the equation $2\cos^2 x + 3\sin x = 3$ for $0^\circ \le x < 360^\circ$.

Mark schemeshow ▾
(a) M1 use $\cos^2 x = 1-\sin^2 x$: $2-2\sin^2 x+3\sin x=3$  A1 rearrange: $2\sin^2 x-3\sin x+1=0$ ∎.
(b) M1 factorise: $(2\sin x-1)(\sin x-1)=0$  A1 $\sin x=\dfrac12$ or $\sin x=1$  A1 $x=30^\circ,\ 150^\circ$  A1 $x=90^\circ$.
Q7. 6 marks

(a) Show that $\dfrac{\sin\theta}{1-\cos\theta} + \dfrac{1-\cos\theta}{\sin\theta} \equiv \dfrac{2}{\sin\theta}$. (b) Hence solve, for $0^\circ < \theta < 360^\circ$, the equation $\dfrac{\sin\theta}{1-\cos\theta} + \dfrac{1-\cos\theta}{\sin\theta} = 4$.

Mark schemeshow ▾
(a) M1 single fraction: $\dfrac{\sin^2\theta+(1-\cos\theta)^2}{(1-\cos\theta)\sin\theta}$  M1 numerator $=\sin^2\theta+1-2\cos\theta+\cos^2\theta = 2-2\cos\theta$ using $\sin^2\theta+\cos^2\theta=1$  A1 $=\dfrac{2(1-\cos\theta)}{(1-\cos\theta)\sin\theta}=\dfrac{2}{\sin\theta}$ ∎.
(b) M1 $\dfrac{2}{\sin\theta}=4 \Rightarrow \sin\theta=\dfrac12$  A1 $\theta=30^\circ$  A1 $\theta=150^\circ$.
Q8. 7 marks

(a) Show that the equation $2\sin^2 x = 1 + \cos x$ can be written in the form $2\cos^2 x + \cos x - 1 = 0$. (b) Hence solve the equation $2\sin^2 x = 1 + \cos x$ for $0^\circ \le x < 360^\circ$, giving all solutions.

Mark schemeshow ▾
(a) M1 use $\sin^2 x = 1-\cos^2 x$: $2-2\cos^2 x = 1+\cos x$  A1 rearrange: $2\cos^2 x+\cos x-1=0$ ∎.
(b) M1 factorise: $(2\cos x-1)(\cos x+1)=0$  A1 $\cos x=\dfrac12$ or $\cos x=-1$  A1 $x=60^\circ$  A1 $x=300^\circ$  A1 $x=180^\circ$.
Q9. 7 marks

Solve, for $0^\circ \le x < 360^\circ$, the equation $\sin(2x-30^\circ) = \dfrac{1}{2}$. Give all solutions, and show clearly how you know your list is complete.

Mark schemeshow ▾
M1 let $u=2x-30^\circ$, widen the interval: $-30^\circ \le u < 690^\circ$  A1 $u=30^\circ$  A1 $u=150^\circ$  M1 add $360^\circ$: $u=390^\circ,\ 510^\circ$  A1 complete list $u=30^\circ,150^\circ,390^\circ,510^\circ$ (next values $750^\circ$, $-210^\circ$ out of range)  A1 $x=30^\circ,\ 90^\circ$  A1 $x=210^\circ,\ 270^\circ$.
Q10. 8 marks

(a) Prove that $\dfrac{1}{1+\sin\theta} + \dfrac{1}{1-\sin\theta} \equiv \dfrac{2}{\cos^2\theta}$. (b) Hence solve, for $0^\circ \le \theta < 360^\circ$, the equation $\dfrac{1}{1+\sin\theta} + \dfrac{1}{1-\sin\theta} = 8$.

Mark schemeshow ▾
(a) M1 common denominator: $\dfrac{(1-\sin\theta)+(1+\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}$  A1 $=\dfrac{2}{1-\sin^2\theta}$  M1 use $\sin^2\theta+\cos^2\theta=1$, so $1-\sin^2\theta=\cos^2\theta$  A1 $=\dfrac{2}{\cos^2\theta}$ ∎.
(b) M1 $\dfrac{2}{\cos^2\theta}=8 \Rightarrow \cos^2\theta=\dfrac14$  A1 $\cos\theta=\pm\dfrac12$  A1 $\theta=60^\circ,\ 300^\circ$  A1 $\theta=120^\circ,\ 240^\circ$.