# 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) . 
(b) Hence or otherwise, solve sin 2x + cos 2x – 1 + cos x – sin x = 0 for 0 < x < 2π . 

Ans  ## Question

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

## 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)

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 $$4 – \cos 2\theta + 5\sin \theta = 2{\sin ^2}\theta + 5\sin \theta + 3$$ .


a.

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


b.

## 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. $$(2\sin \theta + 3)(\sin \theta + 1)$$ , $$(2x + 3)(x + 1) = 0$$ , $$\sin x = \frac{{ – 5 \pm \sqrt 1 }}{4}$$
correct solution $$\sin \theta = – 1$$ (do not penalise for including $$\sin \theta = – \frac{3}{2}$$     (A1)
$$\theta = \frac{{3\pi }}{2}$$     A2     N3