Standard +0.3 This is a straightforward double angle equation requiring the identity tan(2x) = 2tan(x)/(1-tan²(x)), leading to a simple quadratic in tan(x). The solution involves standard algebraic manipulation and finding angles in a specified interval, making it slightly easier than average but requiring more than just recall.
EITHER: Use \(\tan 2A\) formula and obtain a horizontal equation in \(\tan x\)
M1
Simplify the equation to the form \(3\tan^2 x = 1\), or equivalents
A1
Obtain answer \(30°\)
A1
Obtain second answer \(150°\) and no others in the range
A1
OR: Use \(\sin 2A\) and \(\cos 2A\) formulae and obtain a horizontal equation in \(\sin x\) or \(\cos x\)
M1
Simplify the equation to \(4\sin^2 x = 1.4\cos^2 x = 3\), or equivalent
A1
Obtain answer \(30°\)
A1
Obtain second answer \(150°\) and no others in the range
A1
[Ignore answers outside the given range]
[Treat answers in radians as a MR and deduct one mark from the marks for the angles.]
[Methods leading to an equation in \(\cos 3x\) or \(\cos 2x\), or to the equality of two tangents can also earn M1A1, and then A1 + A1 for \(30°\) and \(150°\) only.]
[SR: If the answer \(30°\) is found by inspection or from a graph, and is exactly verified, award B2. If a second answer \(150°\) is found and verified, and no others stated, award B2]
4
EITHER: Use $\tan 2A$ formula and obtain a horizontal equation in $\tan x$ | M1 |
Simplify the equation to the form $3\tan^2 x = 1$, or equivalents | A1 |
Obtain answer $30°$ | A1 |
Obtain second answer $150°$ and no others in the range | A1 |
OR: Use $\sin 2A$ and $\cos 2A$ formulae and obtain a horizontal equation in $\sin x$ or $\cos x$ | M1 |
Simplify the equation to $4\sin^2 x = 1.4\cos^2 x = 3$, or equivalent | A1 |
Obtain answer $30°$ | A1 |
Obtain second answer $150°$ and no others in the range | A1 |
[Ignore answers outside the given range] | | |
[Treat answers in radians as a MR and deduct one mark from the marks for the angles.] | | |
[Methods leading to an equation in $\cos 3x$ or $\cos 2x$, or to the equality of two tangents can also earn M1A1, and then A1 + A1 for $30°$ and $150°$ only.] | | |
[SR: If the answer $30°$ is found by inspection or from a graph, and is exactly verified, award B2. If a second answer $150°$ is found and verified, and no others stated, award B2] | | |
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