Standard +0.8 This question requires converting sec x to 1/cos x, multiplying through by cos x (noting restrictions), rearranging to a quadratic in sin x, and carefully checking solutions against the domain restriction cos x ≠ 0. The multi-step algebraic manipulation, domain considerations, and solving within a specific interval elevate this above a standard trig equation.
divide through by \(\cos x\) to obtain \(2\tan x + \sec^2 x = 4\)
B1
\(2\tan x + \tan^2 x + 1 = 4\)
M1*
use of Pythagoras to obtain equation in \(\tan x\) only; allow 1 sign error
\(\tan^2 x + 2\tan x - 3 [=0]\)
A1
\(\tan x = 1\) or \(-3\)
M1*dep
2 values obtained for \(\tan x\) from their quadratic
\([x =]\ -1.24905\) to \(-1.249\) or \(-1.25\) or \(-1.2\)
\([x =]\ 1.8925\) to \(1.893\) or \(1.89\) or \(1.9\)
A1
any two correct
\([x =]\ \frac{\pi}{4}\) or \(0.785\) to \(0.7854\) or \(0.79\)
\([x =]\ -\frac{3\pi}{4}\) or \(-2.3562\) to \(-2.356\) or \(-2.36\) or \(-2.4\)
A1
all four correct and no extra values in range; ignore correct extra values outside range but A0 if incorrect values outside range
Alternatively: multiply through by \(\cos x\) to obtain \(2\sin x\cos x + 1 = 4\cos^2 x\)
Answer
Marks
Guidance
\(\sin 2x + 1 = 2\cos 2x + 2\)
M1*
use of double angle formulae, allow 1 sign error
\(5\cos^2 2x + 4\cos 2x [=0]\)
A1
or \(\sqrt{5}\cos(2x+0.4636...) = -1\) or \(\sqrt{5}\sin(2x-1.1071...)=1\)
NB square both sides: \(\sin^2 2x = 4\cos^2 2x + 4\cos 2x + 1\)
or \(\cos(2x+0.4636...)= -\frac{1}{\sqrt{5}}\) or \(\sin(2x-1.1071...)=\frac{1}{\sqrt{5}}\)
\(\cos 2x = 0\) or \(-0.8\); 2 values obtained for \(\cos 2x\) from their quadratic
M1dep*
\([x=]\ -1.24905\) to \(-1.249\) or \(-1.25\) or \(-1.2\)
\([x=]\ 1.8925\) to \(1.893\) or \(1.89\) or \(1.9\)
A1
any two correct
\([x=]\ \frac{\pi}{4}\) or \(0.785\) to \(0.7854\) or \(0.79\)
\([x=]\ -\frac{3\pi}{4}\) or \(-2.3562\) to \(-2.356\) or \(-2.36\) or \(-2.4\)
A1
all four correct and no extra values in range; ignore correct extra values outside range but A0 if incorrect values outside range
## Question 17:
divide through by $\cos x$ to obtain $2\tan x + \sec^2 x = 4$ | B1 |
$2\tan x + \tan^2 x + 1 = 4$ | M1* | use of Pythagoras to obtain equation in $\tan x$ only; allow 1 sign error
$\tan^2 x + 2\tan x - 3 [=0]$ | A1 |
$\tan x = 1$ or $-3$ | M1*dep | 2 values obtained for $\tan x$ from their quadratic
$[x =]\ -1.24905$ to $-1.249$ or $-1.25$ or $-1.2$ | |
$[x =]\ 1.8925$ to $1.893$ or $1.89$ or $1.9$ | A1 | any two correct
$[x =]\ \frac{\pi}{4}$ or $0.785$ to $0.7854$ or $0.79$ | |
$[x =]\ -\frac{3\pi}{4}$ or $-2.3562$ to $-2.356$ or $-2.36$ or $-2.4$ | A1 | all four correct and no extra values in range; ignore correct extra values outside range but A0 if incorrect values outside range
**Alternatively:** multiply through by $\cos x$ to obtain $2\sin x\cos x + 1 = 4\cos^2 x$
$\sin 2x + 1 = 2\cos 2x + 2$ | M1* | use of double angle formulae, allow 1 sign error
$5\cos^2 2x + 4\cos 2x [=0]$ | A1 | or $\sqrt{5}\cos(2x+0.4636...) = -1$ or $\sqrt{5}\sin(2x-1.1071...)=1$
**NB** square both sides: $\sin^2 2x = 4\cos^2 2x + 4\cos 2x + 1$ | | or $\cos(2x+0.4636...)= -\frac{1}{\sqrt{5}}$ or $\sin(2x-1.1071...)=\frac{1}{\sqrt{5}}$
$\cos 2x = 0$ or $-0.8$; 2 values obtained for $\cos 2x$ from their quadratic | M1dep* |
$[x=]\ -1.24905$ to $-1.249$ or $-1.25$ or $-1.2$ | |
$[x=]\ 1.8925$ to $1.893$ or $1.89$ or $1.9$ | A1 | any two correct
$[x=]\ \frac{\pi}{4}$ or $0.785$ to $0.7854$ or $0.79$ | |
$[x=]\ -\frac{3\pi}{4}$ or $-2.3562$ to $-2.356$ or $-2.36$ or $-2.4$ | A1 | all four correct and no extra values in range; ignore correct extra values outside range but A0 if incorrect values outside range
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17 In this question you must show detailed reasoning.
Solve the equation $2 \sin x + \sec x = 4 \cos x$, where $- \pi < x < \pi$.
\hfill \mbox{\textit{OCR MEI Paper 2 2023 Q17 [6]}}