Deduce related solution - Standard trigonometric equations

9 marks Standard +0.3

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
1. Show that the equation
$$\sin \theta ( 7 \sin \theta - 4 \cos \theta ) = 4$$ can be written as $$3 \tan ^ { 2 } \theta - 4 \tan \theta - 4 = 0$$
Hence solve, for $0 < x < 360 ^ { \circ }$ $$\sin x ( 7 \sin x - 4 \cos x ) = 4$$ giving your answers to one decimal place.
Hence find the smallest solution of the equation $$\sin 4 \alpha ( 7 \sin 4 \alpha - 4 \cos 4 \alpha ) = 4$$ in the range $720 ^ { \circ } < \alpha < 1080 ^ { \circ }$, giving your answer to one decimal place.

7 marks Moderate -0.3

Given that $$3\sin^2 x - 8\cos x - 7 = 0,$$ show that, for real values of $x$, $$\cos x = -\frac{2}{3}.$$ [3]
Hence solve the equation $$3\sin^2(\theta + 70°) - 8\cos(\theta + 70°) - 7 = 0$$ for $0° \leqslant \theta \leqslant 180°$. [4]

6 marks Moderate -0.8

Given that $$5 \cos \theta - 2 \sin \theta = 0,$$ show that $\tan \theta = 2.5$ [2]
Solve, for $0 \leq x \leq 180$, the equation $$5 \cos 2x° - 2 \sin 2x° = 0,$$ giving your answers to 1 decimal place. [4]

6 marks Moderate -0.3

1. Show that $\cos \theta = \frac{1}{2}$ is one solution of the equation $$6\sin^2 \theta + 5\cos \theta = 7$$ [2 marks]
2. Find all the values of $\theta$ that solve the equation $$6\sin^2 \theta + 5\cos \theta = 7$$ for $0° \leq \theta \leq 360°$ Give your answers to the nearest degree. [2 marks]
Hence, find all the solutions of the equation $$6\sin^2 2\theta + 5\cos 2\theta = 7$$ for $0° \leq \theta \leq 360°$ Give your answers to the nearest degree. [2 marks]

7 marks Standard +0.3

1. Show that the equation $$3\sin\theta\tan\theta = 5\cos\theta - 2$$ is equivalent to the equation $$(4\cos\theta - 3)(2\cos\theta + 1) = 0$$ [3 marks]
2. Solve the equation $$3\sin\theta\tan\theta = 5\cos\theta - 2$$ for $-180° \leq \theta \leq 180°$ [2 marks]
Hence, deduce all the solutions of the equation $$3\sin\left(\frac{1}{2}\theta\right)\tan\left(\frac{1}{2}\theta\right) = 5\cos\left(\frac{1}{2}\theta\right) - 2$$ for $-180° \leq \theta \leq 180°$, giving your answers to the nearest degree. [2 marks]

7 marks Moderate -0.3

1. By using a suitable trigonometric identity, show that the equation $$\sin \theta \tan \theta = 4 \cos \theta$$ can be written as $$\tan^2 \theta = 4$$ [1 mark]
2. Hence solve the equation $$\sin \theta \tan \theta = 4 \cos \theta$$ where $0^\circ < \theta < 360^\circ$ Give your answers to the nearest degree. [3 marks]
Deduce all solutions of the equation $$\sin 3\alpha \tan 3\alpha = 4 \cos 3\alpha$$ where $0^\circ < \alpha < 180^\circ$ Give your answers to the nearest degree. [3 marks]

6 marks Standard +0.3

Solve, for $-180° \leq x < 180°$, the equation $$3 \sin^2 x + \sin x + 8 = 9 \cos^2 x$$ giving your answers to 2 decimal places. [4]
Hence find the smallest positive solution of the equation $$3\sin^2(2\theta - 30°) + \sin(2\theta - 30°) + 8 = 9 \cos^2(2\theta - 30°)$$ giving your answer to 2 decimal places. [2]

8 marks Standard +0.3

Show that the equation $$2 \sin x \tan x = \cos x + 5$$ can be expressed in the form $$3 \cos^2 x + 5 \cos x - 2 = 0.$$ [3]
Hence solve the equation $$2 \sin 2\theta \tan 2\theta = \cos 2\theta + 5,$$ giving all values of $\theta$ between $0°$ and $180°$, correct to $1$ decimal place. [5]

8 marks Standard +0.3

Show that the equation $$4\cos\theta - 1 = 2\sin\theta\tan\theta$$ can be written in the form $$6\cos^2\theta - \cos\theta - 2 = 0$$ [4]
Hence solve, for $0 \leq x < 90°$ $$4\cos 3x - 1 = 2\sin 3x\tan 3x$$ giving your answers, where appropriate, to one decimal place. (Solutions based entirely on graphical or numerical methods are not acceptable.) [4]