can some one please help me wth this. Find all the values of x in the interval 0=<x=<180?

Question

which satisfy the equation sin3x + 2cos3x = 0.
note :0 and 180 are in degrees.
please explain very well.
thank you very much

Answer 1

180=56-1+8

Answer 2

rearrage the equation in steps as follows

1) Divide by cos3x

sin3x/cos3x +2cos3x/cos3x = 0

2) simplify

tan3x + 2 = 0

tax 3x = -2


what this means is that you have a triangle where x is negative and y is twice x. Notice that y is never negative in the range specified, so x must be negative for the tangent to be negative. A 60 degree triangle, sitting in the second quadrent satisfies this relationship. Thus 3X = 180 - 60 =120

or x = 40, and there is only one.

check sin(120) + 2 cos(120) = 0
sqrt(3) + 2 (-sqrt(3)/2) = 0

Answer 3

Let z = 3x and substitute:
sin z + 2 cos z = 0
subtract 2 cos z from both sides:
sin z = - 2 cos z
divide both sides by cos z:
sin z / cos z = -2 = tan z
arc tan –2 = z = -63.4349488229220106484278062795467

For now, we will just round this down to -63 degrees and remember to insert the more accurate value later.

An angle of –63 degrees is a solution in the fourth quadrant if sin z is negative and cos z positive.

If sin z is positive and cos z is negative, the solution is in the second quadrant and z = 180-63 = 117 degrees.

In general, either of these two angles will satisfy sin z / cos z = -2 as will any angle that is one of these two angles plus a multiple (positive or negative) of 180 degrees.

This is best seen graphically by first sketching an x-y orthogonal coordinate system: an horizontal line (x-axis) intersecting a vertical line (y-axis) at ninety degrees. Draw a line through the origin that makes an angle of 63 degrees with the x axis.

When measuring angles on an x-y coordinate system, you start on the +x axis and rotate about the origin in a counter-clockwise direction for positive angles and clockwise for negative angles. The line you just drew extends from the origin into the fourth quadrant. To get there, you can either rotate from the +x axis clockwise to the line, so the angle you rotate through is –63 degrees, or you can rotate counter-clockwise through an angle of (360 – 63) = 297 degrees.

Now extend the line from the origin in the opposite direction so it extends into the second quadrant.

By definition, at ANY point on this line the tangent of the angle the line makes with the x-axis is equal to the y-axis value of the point divided by the x-axis value.

To see this, pick any two arbitrary points on the line, one point in the fourth quadrant and the other point in the second quadrant. From each point draw a vertical line that intersects the horizontal or x-axis. You now have two similar right triangles. By definition, the tangent of the interior angle at the origin is the y-axis value divided by the x-axis value. This interior angle is 63 degrees for both triangles.

In the fourth quadrant, the angle of the line about the origin, measured from the x-axis, is either –63 degrees (measured clock-wise) or 360-63 = 297 degrees (measured counter-clockwise).

In the second quadrant, the angle about the origin, again measured from the +x axis, is either (180-63) = 117 degrees measured counter-clockwise or (-63 –180) = -243 degrees measured clockwise.

Any of these angles is a solution to sin z + 2 cos z = 0, but not all of them satisfy the other condition given.

The triangle in the second quadrant has a negative value for the x-axis length and a positive value for the y-axis length. The ratio of y-axis value (side opposite the interior angle) to the x-axis value (side adjacent to the interior angle) is the tangent of the angle measured either clockwise OR counter-clockwise from the +x axis. This value is –2.

The triangle in the fourth quadrant has a negative value for the y-axis length and a positive value for the x-axis length. Again, the ratio of y-axis value (side opposite the interior angle) to the x-axis value (side adjacent to the interior angle) is the tangent of the angle, measured either clockwise OR counter-clockwise from the +x axis. This value is again –2.

Since we want solutions for 0 =< x <= 180, this means 0 =< z <= 540. By examination (and realizing we can also add multiples of 180 degrees to any solution), the values of z must be 117 degrees, or 297 degrees or 477 degrees. Since z = 3x, the only thing left to do is divide z by three: x = 39 degrees, or 99 degrees or 159 degrees.

Oh yeah, we should go back and substitute z = -63.4349488229220106484278062795467 degrees to get values of z = 116.565051177077989351572193720453 degrees, or 296.565051177077989351572193720453 degrees, or
476.565051177077989351572193720453 degrees.

And after dividing these by three, x = 38.8550170590259964505240645734844 degrees, or
98.8550170590259964505240645734844 degrees, or
158.855017059025996450524064573484 degrees.

If you sketch those little triangle diagrams you will easily be able to determine whether sin or cos or tan of any angle is positive or negative. Just remember the angle is always the acute angle measured with respect to the x-axis no matter what quadrant it ends up in. Also, when calculating sin and cos, the hypotenuse is always positive.

Answer 4

2cos3x= -sin3x...................(1)

=>

(2cos3x)^2=(-sin3x)^2

=>

4 cos^2 3x = sin^2 3x

we add cos^2 3x in both

5cos^2 3x = 1 (because sin^2 3x +cos^2 3x =1)

cos^2 3x = 0.2

cos3x =0.4472

we can get that x is located in 63 to 64 degrees

and we consider (1)

so the answer must 3X= 180-63.** and 360-63.**

(because when x is located in 90 to 180 =>sin(+) cos(-)

and when x is located in 270 to 360 => sin(-) cos(+))



i.e. 3x = 116.** or 296.** and plus 116.**+360

x= 38.** or 98.** or 158.**

Answer 5

So you want to solve

sin(3x) + 2cos(3x) = 0.

Bring the 2cos(3x) over to the right hand side, to get

sin(3x) = -2cos(3x)

Now, divide both sides by cos(3x)

sin(3x) / cos(3x) = -2

And note that sine over cosine is equal to tan.

tan(3x) = -2

At this point, you ask yourself: where on the graph is tan equal to -2? What we DO know is that it's going in be in quadrants 1 and 3; however, since 0 <= x <= 180, that leaves out quadrant 3, so it's going to be in quadrant 1. Unfortunately -2 isn't one of our known values, so we take the arctan of both sides. Remember to add 4 more values just to be safe.

3x = arctan(-2), arctan(-2) + 360, arctan(-2) + 720

x = arctan(-2)/3, arctan(-2)/3 + 120, arctan(-2)/3 + 240

Cross out any values that don't fall between 0 and 180 degrees.

Answer 6

let y=3x
sin y+2cos y=0 0<=y<=540
sin y=-2cos y
sin y/cos y=-2
y=arctan -2=-63.434948822922010648427806279547
since tan y<0, 2nd & 4th quadrants
90<=y<=180
y=180-63.4349=116.5651°
x=116.5651/3=38.8550°
270<=y<=360
y=360-63.4349=296.5651°
x=296.5651/3=98.8550°
450<=y<=540
y=540-63.4369=476.5651°
x=476.5651/3=158.8550°

x=38.8550°
x=98.8550°
x=158.8550°