Standard +0.3 This is a straightforward two-part question requiring standard algebraic manipulation (converting tan to sin/cos, using Pythagorean identity) followed by solving a quadratic in cos and applying a linear transformation to the angle. The techniques are routine for Further Maths students with no novel insight required.
5. (a) Show that the equation
$$4 \cos \theta - 1 = 2 \sin \theta \tan \theta$$
can be written in the form
$$6 \cos ^ { 2 } \theta - \cos \theta - 2 = 0$$
(b) Hence solve, for \(0 \leqslant x < 90 ^ { \circ }\)
$$4 \cos 3 x - 1 = 2 \sin 3 x \tan 3 x$$
giving your answers, where appropriate, to one decimal place.
(Solutions based entirely on graphical or numerical methods are not acceptable.) [0pt]
5. (a) Show that the equation
$$4 \cos \theta - 1 = 2 \sin \theta \tan \theta$$
can be written in the form
$$6 \cos ^ { 2 } \theta - \cos \theta - 2 = 0$$
(b) Hence solve, for $0 \leqslant x < 90 ^ { \circ }$
$$4 \cos 3 x - 1 = 2 \sin 3 x \tan 3 x$$
giving your answers, where appropriate, to one decimal place.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)\\[0pt]
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\hfill \mbox{\textit{SPS SPS FM Pure 2023 Q5 [8]}}