A-LevelMathematicsTrigonometry9709/32 · May/June 2023 · Q45 Marks
Solve the equation 2 cos x – cos (x/2) = 1 for 0 ≤ x ≤ 2π.
- Solve the equation 2 cos x – cos (x/2) = 1 for 0 ≤ x ≤ 2π.[5]
Mark scheme
Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.
- M1 Use correct double angle formula to obtain an equation in cos(x/2) only [e.g. 2(2cos²(x/2)-1)-cos(x/2)=1.]
- A1 Obtain a 3 term quadratic in cos(x/2) [e.g. 4cos²(x/2)-cos(x/2)-3=0. Allow 4cos² u - cosu – 3 = 0 . Condone x/2 = x .]
- A1 Obtain cos(x/2) = -3/4 and cos(x/2) = 1 [Allow answer in u e.g. (4cosu + 3)(cosu -1) and condone x/2=x.]
- DM1 Solve for the original x [Must see evidence of doubling, not halving.]
- A1 Obtain x = 0 and 4.84 and no others in the interval [Ignore any answers outside interval. Accept AWRT 4.84. Accept 1.54π. Must be in radians. 277.2 indicates M1 but is A0.]
- Alternative Method for Question 4:
- M1 Use correct double angle formula to obtain an equation in cosx only [e.g. 2cosx-1 = (cosx+1)/2.]
- A1 Obtain a 3 term quadratic in cosx, [e.g. 8cos²x-9cosx+1=0.]
- A1 Obtain cosx = 1/8 and cos x = 1
- DM1 Solve for x
- A1 Obtain answers x = 0 and 4.84 and no others in the interval [Ignore any answers outside interval. Accept AWRT 4.84. Must be in radians. 277.2 is A0.]
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