Standard +0.3 This question requires using the identity sec²x = 1 + tan²x to convert to a single trig function, then solving a quadratic equation in tan x. While it involves multiple steps (substitution, algebraic manipulation, solving quadratic, finding angles), these are standard techniques for P2 level. The restricted domain and reciprocal functions add minor complexity, but this remains a routine textbook-style question slightly easier than average A-level difficulty.
Use \(\tan^2 x = \sec^2 x - 1\) or \(\sin^2 x = 1 - \cos^2 x\)
M1
Obtain 3-term quadratic in \(\sec x\) or \(\cos x\), e.g. \(2\sec^2 x + \sec x - 6 = 0\)
A1
Make reasonable solution attempt at a 3-term quadratic
M1
Obtain \(\sec x = \frac{3}{2}\) and \(\sec x = -2\), or equivalent
A1
[or \(6\cos^2 x - \cos x - 2 = 0\) and \(\cos x = \frac{2}{3}, -\frac{1}{2}\)]
Obtain answer \(x = 48.2°\)
A1
Obtain answer \(x = 120°\) and no others in the range
A1
[6]
[Ignore answers outside the given range.]
Use $\tan^2 x = \sec^2 x - 1$ or $\sin^2 x = 1 - \cos^2 x$ | M1 |
Obtain 3-term quadratic in $\sec x$ or $\cos x$, e.g. $2\sec^2 x + \sec x - 6 = 0$ | A1 |
Make reasonable solution attempt at a 3-term quadratic | M1 |
Obtain $\sec x = \frac{3}{2}$ and $\sec x = -2$, or equivalent | A1 |
[or $6\cos^2 x - \cos x - 2 = 0$ and $\cos x = \frac{2}{3}, -\frac{1}{2}$] |
Obtain answer $x = 48.2°$ | A1 |
Obtain answer $x = 120°$ and no others in the range | A1 | [6]
[Ignore answers outside the given range.]