Solve the equation for the principal values of x.
Express in degrees.
Chapter 7
2sin2 x + sin x – 1 = 0
(2sin x – 1 )(sin x + 1) = 0
2sin x – 1 = 0 OR sin x + 1 = 0
Section 5
Solving Trigonometric Equations
sin x = 1/2 OR sin x = -1
x = 30 OR
x = -90
Example 2… don’t skip this
Solve the equation for the principal values of x.
Solve the equation for 0  x < 2
Express in degrees.
sin2x  sin x + 1 = cos2 x
3 tan( 2 x)  3  0
3 tan( 2 x)  3
3
tan( 2 x) 
3
 3
2 x  tan 1  
 3 
2 x  30 or 2 x  210
x  15 or x  105
We need to consider
both points on the first
full rotation, because
sometimes when you
finish the algebra all of
your answers are
principle values.
105 is not a principle
value and 15 is
sin2x  sin x + 1 = 1 - sin2 x
sin2x  sin x = - sin2 x
Notice that the
domain is limited
hint: rewrite as 1
trig function
Pythagorean identity
2sin2x  sin x = 0
factor
sin x(2sinx  1) = 0
sin x = 0 or 2sinx  1 = 0
set each factor = 0
sin x = 0
or sinx = 1/2
x = 0, , /6, 5/6
Answer x = 15
1
Solve the equation for all real values of x.
4 cos x sin x  2 sin x  2 3 cos x  3  0
(2 sin x  3)(2 cos x  1)  0
2 sin x  3  0 or 2 cos x  1  0
sin x 
3
2
or
1
cos x 
2
These are in the
These are in the
1st and 2nd quadrants
1st and 4th quadrants
x

3
 2k ,
5
2
 2k
x
 2k , x 
3
3
Hint: FACTOR
Solve the inequality for all real values of 0   < 2.
2cos  +1 < 0
2
 1 3 
 , ,  for  
3
 2 2 
2cos  < -1
cos 
1
2
4
 1  3 
 ,
 for  
3
 2 2 
Refer to unit circle we need x-coordinates less than -1/2
2
4
 
3
3
Turn to page 458
10 minutes with a buddy
2