Solution of Past Paper of CAIE Math 9709/12 Feb/Mar 2020 Question # 11

CIE/GCE/AS/Math/P1/20/Mar/12/Q#11

Question

i. Solve the equation 3 tan2 𝑥 − 5 tan 𝑥 − 2 = 0 for 0𝑜 ≤ 𝑥 ≤ 180𝑜 .
ii. Find the set of values of 𝑘 for which the equation 3 tan2 𝑥 − 5 tan 𝑥 + 𝑘 = 0 has
no solution.
iii. For the equation 3 tan2 𝑥 − 5 tan 𝑥 + 𝑘 = 0, state the value of 𝑘 for which there
are three solutions in the interval 0𝑜 ≤ 𝑥 ≤ 180𝑜 , and find these solutions.


Solution

i.


We have the equation;
3 tan2 𝑥 − 5 tan 𝑥 − 2 = 0
Let tan 𝑥 = 𝑥;
3𝑥 2 − 5𝑥 − 2 = 0


3𝑥 2 − 6𝑥 + 𝑥 − 2 = 0


3𝑥(𝑥 − 2) + 1(𝑥 − 2) = 0


(3𝑥 + 1)(𝑥 − 2) = 0
Now we have two options.


3𝑥 + 1 = 0 𝑥−2=0

3𝑥 = −1 𝑥=2
1
𝑥=−
3

, CIE/GCE/AS/Math/P1/20/Mar/12/Q#11


Since tan 𝑥 = 𝑥;


1
tan 𝑥 = − tan 𝑥 = 2
3
1
𝑥 = tan−1 (− ) 𝑥 = tan−1 (2)
3
Using calculator;

𝑥 = −18.43𝑜 𝑥 = 63.43𝑜




Properties of 𝐭𝐚𝐧 𝜽
𝜋
Domain {𝜃|𝜃 ≠ + 𝑘𝜋, 𝑘 =, … , −1,0,1, … , }
2
Range −∞ ≤ tan 𝜃 ≤ +∞

tan(𝜃 ± 𝑛𝜋) = tan 𝜃
Periodicity
tan(𝜃 ± 𝑛180𝑜 ) = tan 𝜃

Odd/Even tan(−𝜃) = − tan 𝜃

Translation/ tan(𝜋 − 𝜃) = − tan 𝜃
Symmetry tan(180𝑜 − 𝜃) = − tan 𝜃


We utilize the periodicity property of tan 𝜃 to find other solutions (roots) of tan 𝜃: .

Therefore;
tan(𝜃 ± 𝑛180𝑜 ) = tan 𝜃

𝜃 = 𝜃 ± 𝑛180𝑜