8.5 SOLVING MORE
DIFFICULT
TRIGONOMETRIC
EQUATIONS
SOME EQUATIONS ARE IN A QUADRATIC
FORM
2 cos x  7 cos x  3  0
2
 This looks a lot like ax 2 +bx+c=0
 Factor by grouping
2 cos x  6 cos x  1 cos x  3  0
2
(2 cos x  6 cos x )  (1 cos x  3)  0
2
2 cos x (cos x  3)  1(cos x  3)  0
(2 cos x  1)(cos x  3)  0
cos x 
1
2
; cos x  3
cos x 
1
; cos x  3
2
Now determine everywhere that cos(x)=1/2.
x =
𝝅
𝟑
+ 𝟐𝒌𝝅 , x =
𝟓𝝅
𝟑
+ 𝟐𝒌𝝅
We want to do the same thing for cos(x)=3,
but in this case the cosine function never
gets larger than 1, so there is no solution.
EVERY ONCE AND A WHILE, A TRIG
IDENTIT Y NEEDS TO BE APPLIED FIRST
1  sin x  2 cos x
2
 This is dif ficult to solve because of the sine and cosine in the equations
1  sin x  2(1  sin x )
2
 Now set equal to zero, and factor.
sin x tan x  3 sin x
 Subtract 3sinx from both sides.
 Factor out sin(x).
2 cos x  7 cos x  3  0
2
cos x  1  sin x
 Nothing will work here until you square both sides.
2 cos x  1  0
2
2 cos x  sin x  1
2
PG. 326 4-8, 11-14, 19-22