sinx+cos(2x)=0?

Question

Hi
I need to solve this trigonometry problem to find the x-intercepts of a graph between x=0 and x=2pi. I see on the graph that there have to be 3 x-intercepts, but I can only find one answer to this problem. I got x= pi/2. My computer tells me the other two have to be x=3.66519 and x=5.75959 but I don't know how to get those answers. Can you help me with that?

Answer 1

cos(2x)= cos^2x-sin^2x and cos^2x = 1-sin^2x

so sin x +1 -2sin^2x=0 Put sinx =z

2z^2-z-1=0 z= (1+-3)/4 so z= 1 or z=-1/2 sinx =1 x=pi/2
sinx = -1/2 x= pi +pi/6 =7pi/6 and x= 2pi-pi/6= 11pi/6
The first is aprox 3.665 and the second 5.7596

Answer 2

this is my answer to the equation:

sinx+cos(2x)=0 ---> cos(2x) = cos^2(x) - sin^2(x)
= 1 - 2sin^2(x)
sinx + 1 - 2sin^2(x)= 0
2sin^2(x) -sinx - 1 = 0
2 1
1 -1
(2sinx+1)(sinx-1)

sol 1
sinx=1--->x = pi/2
sol2
sinx = -1/2---> x=5pi/4
---->x=7pi/4

i posted a picture, there you will see the other answers the computer gave you, but mine are in radians (the pi stuff)

http://img442.imageshack.us/img442/4486/graph8bx.png

i believe this will be as useful for you as tricky for me...
good luck
nexus

Answer 3

Identity cos 2x = 1 - 2(sin x)^2
Then you have
sin x + [1 - 2(sin x)^2] = 0
Let t = sin x, then
t + 1 - 2t^2 = 0
2t^2 - t - 1 = 0
t = (--1 +- sqr[(-1)^2 - 4(2)(-1)])/2(2) = (1 +- sqr[1+8])/4
= (1 +- 3)/4 = 4/4, -2/4 = 1, -1/2
t = sin x = 1, -1/2