Everything Maths Grade 11

(Marvins-Underground-K-12) #1

17.4 CHAPTER 17. TRIGONOMETRY


cos x =− 0 ,5 [60◦]
II : x = 180◦− 60 ◦+ (360◦. n);n∈Z
= 120◦+ (360◦. n);n∈Z
III : x = 180◦+ 60◦(360◦. n);n∈Z
= 240◦+ (360◦. n);n∈Z

cos x = 1 [90◦]
I;IV : x = 0◦(360◦. n);n∈Z
= (360◦. n);n∈Z

Now we find the specific solutions in the interval [− 180 ◦;360◦]. Appropri-
ate values of n yield

x =− 120 ◦;0◦;120◦;240◦;360◦

Example 13:


QUESTION


Solve for x in the interval [− 360 ◦;360◦]:

2sin^2 x− sin xcos x = 0

SOLUTION


Step 1 : Factorise
Factorising yields

sin x(2sin x− cos x) = 0

which gives two equations

sin x = 0 2sin−cos x = 0
2sin x = cos x
2sin x
cos x

=


cos x
cos x
2tan x = 1
tan x =^12

Step 2 : Solve the two trigonometric equations
General solution:

sin x = 0 [0◦]
∴ x = (180◦. n);n∈Z

tan x =^12 [26, 57 ◦]
I;III : x = 26, 57 ◦+ (180◦. n);n∈Z

Specific solution in theinterval [− 360 ◦;360◦]:

x =− 360 ◦;− 206 , 57 ◦;− 180 ◦;− 26 , 57 ◦;0◦;26, 57 ◦;180◦;206, 25 ◦;360◦
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