" 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.

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

x = 5.339836909/2 = 2.669918455 -----answer.

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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.