IB Math Analysis & Approaches Questionbank-Topic: SL 3.8 Equations leading to quadratic equations in sinx , cosx or tanx SL Paper 1

Question

(a) Show that sin 2x + cos 2x – 1 = 2 sin x (cos x – sin x) . [2]
(b) Hence or otherwise, solve sin 2x + cos 2x – 1 + cos x – sin x = 0 for 0 < x < 2π . [6]

Answer/Explanation

Ans

Question

Solve \(\cos 2x – 3\cos x – 3 – {\cos ^2}x = {\sin ^2}x\) , for \(0 \le x \le 2\pi \) .

Answer/Explanation

Markscheme

evidence of substituting for \(\cos 2x\)     (M1)

evidence of substituting into \({\sin ^2}x + {\cos ^2}x = 1\)     (M1)

correct equation in terms of \(\cos x\) (seen anywhere)     A1

e.g. \(2{\cos ^2}x – 1 – 3\cos x – 3 = 1\) , \(2{\cos ^2}x – 3\cos x – 5 = 0\)

evidence of appropriate approach to solve     (M1)

e.g. factorizing, quadratic formula

appropriate working     A1

e.g. \((2\cos x – 5)(\cos x + 1) = 0\) , \((2x – 5)(x + 1)\) , \(\cos x = \frac{{3 \pm \sqrt {49}}}{4}\)

correct solutions to the equation

e.g. \(\cos x = \frac{5}{2}\) , \(\cos x = – 1\) , \(x = \frac{5}{2}\) , \(x = – 1\)     (A1)

\(x = \pi \)     A1     N4

[7 marks]

Question

Show that \(4 – \cos 2\theta + 5\sin \theta = 2{\sin ^2}\theta + 5\sin \theta + 3\) .

[2]
a.

Hence, solve the equation \(4 – \cos 2\theta + 5\sin \theta = 0\) for \(0 \le \theta \le 2\pi \) .

[5]
b.
Answer/Explanation

Markscheme

attempt to substitute \(1 – 2{\sin ^2}\theta \) for \(\cos 2\theta \)     (M1)

correct substitution     A1

e.g. \(4 – (1 – 2{\sin ^2}\theta ) + 5\sin\theta \)

\(4 – \cos 2\theta + 5\sin \theta = 2{\sin ^2}\theta + 5\sin\theta + 3\)     AG     N0

[2 marks]

a.

evidence of appropriate approach to solve     (M1)

e.g. factorizing, quadratic formula

correct working     A1

e.g. \((2\sin \theta + 3)(\sin \theta + 1)\) , \((2x + 3)(x + 1) = 0\) , \(\sin x = \frac{{ – 5 \pm \sqrt 1 }}{4}\)

correct solution \(\sin \theta = – 1\) (do not penalise for including \(\sin \theta = – \frac{3}{2}\)     (A1)

\(\theta = \frac{{3\pi }}{2}\)     A2     N3

[5 marks]

b.

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