CAIE P1 2016 November — Question 3 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2016
SessionNovember
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicStandard trigonometric equations
TypeQuadratic in sin²/cos²/tan²
DifficultyModerate -0.8 This is a straightforward trigonometric equation requiring standard algebraic manipulation (collecting like terms to get 4sin²x = 6cos²x, then tan²x = 3/2) followed by routine application of inverse trig and quadrant analysis. The technique is direct with no conceptual obstacles, making it easier than average but not trivial due to the algebraic setup and multiple solutions required.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05o Trigonometric equations: solve in given intervals

Showing all necessary working, solve the equation \(6\sin^2 x - 5\cos^2 x = 2\sin^2 x + \cos^2 x\) for \(0° \leq x \leq 360°\). [4]

Question 3:
AnswerMarks
36
4sin2x=6cos2x⇒tan2x= or 4sin2x=6 ( 1−sin2x )
4
[tan x = (±)1.225 or sin x = (±)0.7746 or cos x = (±)0.6325]
x= 50.8 (Allow 0.886 (rad))
Another angle correct
x=50.8°, 1 29.2°, 230.8°, 309.2°
AnswerMarks
[ 0.886, 2.25/6, 4.03, 5.40 (rad) ]M1
A1
A1
AnswerMarks Guidance
A1[4] Or 4 ( 1−cos2x ) =6cos2x
Or any other angle correct
Ft from 1st angle (Allow radians)
All 4 angles correct in degrees
Question 3:
3 | 6
4sin2x=6cos2x⇒tan2x= or 4sin2x=6 ( 1−sin2x )
4
[tan x = (±)1.225 or sin x = (±)0.7746 or cos x = (±)0.6325]
x= 50.8 (Allow 0.886 (rad))
Another angle correct
x=50.8°, 1 29.2°, 230.8°, 309.2°
[ 0.886, 2.25/6, 4.03, 5.40 (rad) ] | M1
A1
A1
A1 | [4] | Or 4 ( 1−cos2x ) =6cos2x
Or any other angle correct
Ft from 1st angle (Allow radians)
All 4 angles correct in degrees
Showing all necessary working, solve the equation $6\sin^2 x - 5\cos^2 x = 2\sin^2 x + \cos^2 x$ for $0° \leq x \leq 360°$. [4]

\hfill \mbox{\textit{CAIE P1 2016 Q3 [4]}}