Can anyone help me on this math problem involving Trig Equations? I know the answer, but how do you to it?

Question

1. 2sinxcosx+cosx=0
The answer is: pi/2, 7pi/6, 3pi/2, 11pi/6....How do you get each of these answers?


2. sinxcosx+cosx=0
answer is: pi/2, 3pi/2

Answer 1

u can factor it out to be (2sinx+1)(cosx)=0.. then find x for which either sinx=-1/2 or cosx=0

(sinx+1)cosx=0 same method applies

Answer 2

1.)
2sinxcosx + cosx = 0
(cosx)(2sinx + 1) = 0
so either cosx = 0 or 2sinx + 1 = 0

2sinx + 1 = 0
2sinx = -1
sinx = (-1/2)
x = sin^-1(-1/2)
x = 330° which is (11pi/6)

cosx = 0
x = cos^-1(0)
x = 90° which is (pi/2) or 270° which is (3pi/2)

So the answers are (pi/2), (3pi/2), and (11pi/2)

I don't think (7pi/6) is correct, since it doesn't give you (-1/2) in sin(x) = (-1/2), or 0 in cosx = 0.

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2.)
sinxcosx + cosx = 0
cosx(sinx + 1) = 0

so cosx = 0 or sinx + 1 = 0

sinx + 1 = 0
sinx = -1
x = sin^-1(-1)
x = 270° which is (3pi/2)

cosx = 0
x = cos^-1(0)
x = 90° which is (pi/2) or 270° which is (3pi/2)

Since you already have 270° for the top

x = (pi/2) or (3pi/2)

www.quickmath.com for #1, will give you (pi/2), (-pi/2), and (-pi/6), which is 90°, -90°, -30°, or 90°, 270°, 330°, which is (pi/2), (3pi/2), and (11pi/6). For #2, it will give you (pi/2) and (-pi/2), which is 90° or 270°, or (pi/2) and (3pi/2)

Also to make it easier for you to convert to degrees, just think of pi as 180°, so when you have (3pi/2), thats ((3 * 180)/2) which gives you (540/2), which gives you 270

Answer 3

#1: (2 sin x + 1)(cos x)=0
2 sin x + 1=0 or cos x=0
sin x = -1/2 or cos x=0

If you consider the unit circle, cos would be the x axis, so cos x=0 at the top and bottom of the circle, that is at ±pi/2, but since your teacher apparently wants an answer between 0 and 2pi, this would be pi/2 and 3pi/2. sin x is the y axis of the unit circle, and equals -1/2 in two places: in the lower right at -pi/6 (-30 degrees - remember your special triangles), and in the lower left left of the circle, at -5pi/6. But again, since your teacher wants this expressed in angles between 0 and 2pi, this is 11pi/6 and 7pi/6, respectively. (If you're wondering, you can just add any integral multiple of ±2pi to an angle and end up at the same point on the circle). Thus your solutions are pi/2, 3pi/2, 7pi/6, and 11pi/6, which is all the answers listed, thus the problem is complete.

#2:(sin x +1)(cos x)=0
sin x + 1=0 or cos x=0
sin x=-1 or cos x=0

Again, the solutions for cos x=0 are pi/2 and 3pi/2, and sin x=-1 only at 3pi/2, which we've already listed, thus your only two solutions are pi/2 and 3pi/2.

Answer 4

2sinxcosx+cosx = 0
cosx (2sinx + 1) = 0
cosx = 0 or 2sinx + 1 = 0
cosx = 0 or 2sinx = -1
cosx = 0 or sinx = -1/2

cosx = 0 will give you x= pi/2
Adding multiples of pi you get all solutions:
pi/2, pi/2 + pi, pi/2 + 2pi, ......
pi/2, 3pi/2, 5pi/2, ......

sinx = -1/2 will give you sin (-x) = 1/2,
so -x = pi/6 or -x = 5pi/6
x = -pi/6 or -5pi/6
Adding multiples of 2pi you will get all the sloutions:
-pi/6, -pi/6 + 2pi, -pi/6 + 4pi, ....
-pi/6, 11pi/6, 23pi/6, ....

and
-5pi/6, -5pi/6 + 2pi, -5pi/6 + 4pi, ....
-5pi/6, 7pi/6, 19pi/6, ....

Your second question can also be done precisely in the same way. Find one solution and then add multiples of 2pi.

Answer 5

1. 2sinxcosx+cosx=0 => 2sinxcosx= -cosx =>
2sinx= -cosx/cosx/2 => sinx= -1/2 then you find the quadrants where sinx= -1/2

2. You do this one the same way.

sinxcosx+cosx=0 => sinx=cosx/cosx => six =1 then you find the quadrants where sixx=1

Answer 6

1.
cos x ( 2sin x + 1 ) = 0
cos x = 0 or 2sin x + 1 = 0
x = pi / 2, x = 7 pi / 6
3 pi / 2, x = 11 pi / 6
etc. etc.

2.
cos x (sin x + 1 ) = 0
cos x = 0 Or sin x + 1 = 0
x = pi / 2, x = 3 pi / 2
3 pi / 2, x = 7 pi / 2
etc. etc.