help me solve : sin2xcosx + cos2xsinx = 1 for x as element between 0 and 2pi?

Question

Thanks!

Answer 1

The first answer starts well, but is incomplete:
sin 3x = 1, x between 0 and 2pi => 3x between 0 and 6pi.
So 3x is pi/2, 5pi/2 or 9pi/2 and hence x is pi/6, 5pi/6 or 3pi/2.

Answer 2

sin(2x)cos(x) + cos(2x)sin(x) = 1

First, use the double angle identities to convert

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

[2sin(x)cos(x)]cos(x) + [1 - 2sin^2(x)]sin(x) = 1

Expand and simplify.

2sin(x)cos^2(x) + sin(x) - 2sin^3(x) = 1

Our desire here is to change everything to all sines. Convert cos^2(x) into 1 - sin^2(x).

2sin(x)[1 - sin^2(x)] + sin(x) - 2sin^3(x) = 1

And expand, to get

2sin(x) - 2sin^3(x) + sin(x) - 2sin^3(x) = 1

Simplifying even more, we get

3sin(x) - 4sin^3(x) = 1

And moving the 1 to the left hand side, we get

3sin(x) - 4sin^3(x) - 1 = 0

Now, let's reorder the equation in descending powers of sine.

-4sin^3(x) + 3sin(x) - 1 = 0

And, for purposes of math etiquette, let's make the term in front of sin^3(x) positive by multiplying both sides by -1.

4sin^3(x) - 3sin(x) + 1 = 0

At this point, this is actually a cubic in disguise. If you don't believe me, let u = sin(x). Then we have

4u^3 - 3u + 1 = 0

Now, we can solve this like we normally solve cubics; we defined a function p(u) = 4u^3 - 3u + 1, and plug in factors of 1 (-1, 1).

Test p(-1): 4(-1)^3 - 3(-1) + 1 = -4 + 3 + 1 = 0.
Since -1 is a root, (u + 1) is a factor.

Since we KNOW (u + 1) is a factor, we can actually use long division to solve the cubic; that is, (u + 1) INTO (4u^3 - 3u + 1).
I won't show you the gory details of the long division, but the answer you should get is (4u^2 - 4u + 1) after the division. This means we get the factorization (u + 1) (4u^2 - 4u + 1). So,

(u + 1) (4u^2 - 4u + 1) = 0

Which factors further into
(u + 1) (2u - 1)^2 = 0

Now, we can equate each factor to 0.
u + 1 = 0, or (2u - 1)^2 = 0. Thus,
u = -1, or 2u - 1 = 0.

u = -1 or u = 1/2

BUT we let u = sin(x). Therefore, what we really have are the two equation

sin(x) = -1 and sin(x) = 1/2

Where on the graph is sin(x) equal to -1? The answer to that is 3pi/2.
Where on the graph is sin(x) equal to 1/2? The answer to that is pi/6 and 5pi/6.

Therefore, x = {3pi/2, pi/6, 5pi/6}

Edit: Made a mistake somewhere in my calculation; It's only partially correct.

Answer 3

sin2xcosx+cos2xsinx=sin(2x+x)
=sin3x
sin3x=sinpi/2
3x=pi/2
x=pi/6
since sin is also poeiive in the second quadrant
xis also 5pi/6
so x=pi/6 or 5pi/6

Answer 4

sin2x cosx + cos2x sinx = sin(2x + x)
= sin3x

If sin3x = 1, then
3x = Pi/2 + 2nPi, n = 0,1,2, . . .
x = Pi/6 + 2nPi/3, n = 0,1,2, . . .

Thus, x = Pi/6, 5Pi/6, 3Pi/2,