# sin and cos

• Aug 28th 2005, 10:11 PM
yesdnil
sin and cos
i can't figure out how to do this problem....
-4+11 sin x = 6 cos^2 x

solve for x....
• Aug 28th 2005, 10:20 PM
ticbol
What do you want to do? Rearrange the equation? Solve for x? What?
• Aug 29th 2005, 12:02 AM
ticbol
So it is solve for x.

-4+11 sin x = 6 cos^2 x

-4 +11sinX = 6cos^2(X) ------(i)

There are two variables in (i). But to the beauty of Trigonometry, by trig identities, we can express one in terms of the other.
It is easier to express cosX in terms of sinX in this case.

Since sin^2(X) +cos^2(X) = 1, then,
cos^2(X) = 1 -sin^2(X)
Substitute that into (i),
-4 +11sinX = 6[1 -sin^2(X)]
-4 +11sinX = 6 -6sin^2(X)]
6sin^2(X) +11sinX -4 -6 = 0
6sin^2(X) +11sinX -10 = 0 ----------(ii)

That (ii) is a quadratic equation in sinX.
Solve for sinX using the Quadratic Formula.
Or factor it if it can be factored---and you know how to

factor it.

Factoring (ii),
(3sinX -2)(2sinX +5) = 0

3sinX -2 = 0
3sinX = 2
sinX = 2/3 = 0.6666667 ------positive.
X = arcsin(0.666667) = 41.81 degrees, in the 1st quadrant
or X = (180 -41.81) = 138.19 deg, in the 2nd quadrant.

That is, sine is positive in the 1st and 2nd quadrants.

Therefore, X = 41.81 or 138.18 deg ------answer.

2sinX +5 = 0
2sinX = -5
sinX = -5/2 = -2.5 ------cannot be because minimum value of sine of any angle is -1.0 only.

--------------
Another way, by using the Quadratic Formula in (ii).

sinX = {-11 +,-sqrt[(11)^2 -4(6)(-10)]} / (2*6)
sinX = {-11 +,-sqrt[361]} / (12)
sinX = {-11 +,-19} / 12

sinX = {-11 +19}/12 = 8/12 = 2/3 = 0.666667
etc....(see above for continuation)....

sinX = {-11 -19}/12 = -30/12 = -10/4 = -2.5
etc...