Moderate -0.8 This is a straightforward trigonometric equation requiring only rearrangement to tan(3x) = -1/2, then solving for x by dividing the general solution by 3. The restricted interval and multiple angle make it slightly more involved than the most basic trig equations, but it remains a routine textbook exercise with no conceptual challenges.
2 Find all the values of \(x\) in the interval \(0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }\) which satisfy the equation \(\sin 3 x + 2 \cos 3 x = 0\).
Use of \(\tan = \sin \div \cos\) with 3x. Co. For 60 + "his". For 120 + "his" and no others in range (ignore excess ans. outside range). Loses last A mark if excess answers in the range. NB. \(\sin^2 3x + \cos^2 3x = 0\) etc. M0. But \(\sin^2 3x = (-2\cos 3x)^2\) plus use of \(s^2 + c^2 = 1\) is OK. Alt. \(\sqrt{5}\sin(3x + \alpha)\) or \(\sqrt{5}\cos(3x - \alpha)\) both OK
$\sin 3x + 2\cos 3x = 0$; $\tan 3x = -2$; $x = 38.9$ (8); $x = 98.9$ (8); $x = 158.9$ (8) | M1, A1, A1∨, A1∨ [4] | Use of $\tan = \sin \div \cos$ with 3x. Co. For 60 + "his". For 120 + "his" and no others in range (ignore excess ans. outside range). Loses last A mark if excess answers in the range. NB. $\sin^2 3x + \cos^2 3x = 0$ etc. M0. But $\sin^2 3x = (-2\cos 3x)^2$ plus use of $s^2 + c^2 = 1$ is OK. Alt. $\sqrt{5}\sin(3x + \alpha)$ or $\sqrt{5}\cos(3x - \alpha)$ both OK
2 Find all the values of $x$ in the interval $0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }$ which satisfy the equation $\sin 3 x + 2 \cos 3 x = 0$.
\hfill \mbox{\textit{CAIE P1 2003 Q2 [4]}}