Thread: Trigonometry Help

Hi I am just sooo confused and I don't know what to do... or even start Can someone please explain this.

The smallest positive number for which

3sin(2x-6)=1

is X= ______


and

3sin(2x-5)=1


and

4cos^2 x-9cosx+2=0


I really don't know how to do these.

" The smallest positive number for which
3sin(2x-6)=1
is X= ______

and
3sin(2x-5)=1

and
4cos^2 x-9cosx+2=0 "

x = a number?
x is part of an angle or is an angle. Angles are in degrees and radians. Degrees are not "numbers", radians are "numbers", so we solve for x in radians.

3sin(2x -6) = 1
sin(2x -6) = 1/3 --------positive sine, so angle (2x -6) is in the 1st or 2nd quadrants.
2x -6 = arcsin(1/3)
2x -6 = 0.339836909 in the 1st quadrant, or (pi -0.339836909) = 2.801755744 in the 2nd quadrant.

2x -6 = 0.339836909
2x = 0.339836909 +6
x = 6.339836909/2 = 3.169918455

2x -6 = 2.801755744
2x = 8.801755744
x = 4.400877872 ----larger than 3.169918455

Therefore, x = 3.169918455 is the smallest number to make 3sin(2x-6)=1 true. ----answer.

---------------------
For 3sin(2x -5) = 1, follow the above and you should arrive at
x = 5.339836909/2 = 2.669918455 -----answer.

--------------------
4cos^2(x) -9cos(x) +2 = 0
That is a quadratic equation in cos(x). For simplicity, let y = cos(x),
4y^2 -9y +2 = 0
(4y -1)(y -2) = 0

4y -1 = 0
y = 1/4
Or, cos(x) = 1/4 ---positive cosine, so angle x is in the 1st or 4th quadrants.
x = arccos(1/4) = 1.318116072 in the 1st quadrant, or (2pi -1.318116072) = 4.965069235 in the 4th quadrant.

y -2 = 0
y = 2
or, cos(x) = 2 ----reject this because maximum value of cosine of any angle is 1.0 only.

Therefore, x = 1.318116702 is smallest number to make 4cos^2(x) -9cos(x) +2 = 0 true. ----answer.

Thank you for the help