pre-calculus help...?

Question

i have to solve for x and i'm not sure how...
4cos(^2)x + 5sinx + 3 = 0

is there some sort of identity i'm supposed to use???

thanks!

Answer 1

Hi,

Your other answer is correct in its logic. You can solve it that way,

However, if you are allowed to use your graphing calculator,
enter Y1 = (cos(x))² + 5sin(x) + 3. If your angles are in degrees, set your window to xmin=0, xmax=360,xscl=45, ymin=-10,ymax=10,yscl=1.

Since you want 4cos²(x) + 5sinx + 3 = 0. look for where the graph crosses the x axis. you can use the CALC zero command to find the 2 values. They are 236.93871 and 303.06129 degrees.

I hope that helps!! :-)

On paper, you would do:

4(1 - sin²(x)) + 5sin(x) + 3 = 0
4 - 4sin²(x) + 5sin(x) + 3 = 0
4sin²(x) - 5sin(x) - 7 = 0
............__________
-(-5) ±√(-5)²-4*4*-7
---------------------------- = sin(x)
............2(4)

........_______
5 ±√25 + 112
------------------ = sin(x)
.......8

........___
5 ±√137
------------ = sin(x)
.......8

5 ±11.7
------------ = sin(x)
....8

Since -1 < sin (x) < 1, you can eliminate the + option in the problem. It's too large.

(5 - 11.7)/8 = -.8375 = sin(x)

When you find sin^(-1) (.8375), then x = -56.9°. This shows the reference angle for this problem is 56.9° in either the third or fourth quadrant where the sine value is negative. 180 + 56.9 = 236.9° and 360 - 56.9 = 303.1°

Answer 2

Use cos^2x = 1-sin^2x
4(1-sin^2x+5sinx +3 = 0
4- 4sin^2x +5sinx + 3= 0
-4sin^2x +5sinx +7 = 0
You now have a quadratic equation in sin x
Use quadratic formula:
sinx = [-5 +/- sqrt(5^2-4(-4)(7))](2(-4))


You should be able to solve it from here.