Solving X using Double Angle Formula

[grapz]

Solve for X:

Sin2x cos x + sin^2 x = 1

This is what i did:

2 sinx cos^2 x + ( 1 - cos^2x ) = 1
2 sinx = 1

sinx= 1/2 x = pi/6, 5/6 pi

However i am only able to solve for one root, there is another and i can't find it


Another question :

Solve for X

Sin 2 x - cos 2 x = 0

Sin 2x - ( 1 - 2sin^2x ) = 0
sin 2x - 1 + 2sin^2x = 0

i am not sure how to factor this


Thanks

 

[morson]

For 1.:

The equation reads something like "Something + \(\displaystyle (sinx)^2\) = 1." You should know that \(\displaystyle (cosx)^2 + (sinx)^2 = 1\), so what does that tell you about the first term on the left side? If you don't understand that, here's another way:

\(\displaystyle cos(x)sin(2x) + (sinx)^2 = 1\)

\(\displaystyle 2cos(x)sin(x)cos(x) + (sinx)^2 = 1\), using double angle identity

\(\displaystyle 2sin(x)(cosx)^2 + 1 - (cosx)^2 = 1\), using another identity

Hence, \(\displaystyle 2sin(x)(cosx)^2 - (cosx)^2 = 0\)

Factorising: \(\displaystyle (cosx)^2*[2sinx - 1] = 0\)

For 2.: \(\displaystyle sin(2x) - cos(2x) = 0\)

Rearranging, \(\displaystyle sin(2x) = cos(2x)\)

Now, \(\displaystyle \frac{sin(2x)}{cos(2x)}\ = 1, cos(2x) \not=\ 0\)

Therefore: \(\displaystyle tan(u) = 1\), where u = 2x