Solution of Past Paper of CAIE Math 9709/13 Oct/Nov 2019 Question # 7

CIE/GCE/AS/Math/P1/19/Nov/13/Q#7

Question


i. Show that the equation 3 cos 4 𝜃 + 4 sin2 𝜃 − 3 = 0 can be expressed as


3𝑥 2 − 4𝑥 + 1 = 0
Where 𝑥 = cos 2 𝜃


ii. Hence solve the equation 3 cos 4 𝜃 + 4 sin2 𝜃 − 3 = 0 for 0𝑜 ≤ 𝜃 ≤ 180𝑜.


Solution


i.


We are given the equation;
3 cos 4 𝜃 + 4 sin2 𝜃 − 3 = 0


We have the trigonometric identity;
sin2 𝜃 + cos 2 𝜃 = 1
From this we can substitute sin2 𝜃 = 1 − cos 2 𝜃 in above equation;


3 cos 4 𝜃 + 4(1 − cos 2 𝜃 ) − 3 = 0


3 cos 4 𝜃 + 4 − 4 cos 2 𝜃 − 3 = 0


3 cos 4 𝜃 − 4 cos 2 𝜃 + 1 = 0


Let 𝑥 = cos 2 𝜃;
3𝑥 2 − 4𝑥 + 1 = 0

, CIE/GCE/AS/Math/P1/19/Nov/13/Q#7


ii.


We are required to solve the equation
3 cos 4 𝜃 + 4 sin2 𝜃 − 3 = 0
for 0𝑜 ≤ 𝜃 ≤ 180𝑜.


From (i) we know that given equation can be written as;


3 cos 4 𝜃 − 4 cos 2 𝜃 + 1 = 0


Let 𝑥 = cos 2 𝜃;
3𝑥 2 − 4𝑥 + 1 = 0


3𝑥2 − 3𝑥 − 𝑥 + 1 = 0


3𝑥(𝑥 − 1) − 1(𝑥 − 1) = 0


(3𝑥 − 1)(𝑥 − 1) = 0
Now we have two options.

3𝑥 − 1 = 0 𝑥−1 =0

3𝑥 = 1 𝑥=1

1
𝑥=
3