Solving X using Double Angle Formula

grapz

Junior Member
Joined
Jan 13, 2007
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80
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
 
For 1.:

The equation reads something like "Something + (sinx)2\displaystyle (sinx)^2 = 1." You should know that (cosx)2+(sinx)2=1\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:

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

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

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

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

Factorising: (cosx)2[2sinx1]=0\displaystyle (cosx)^2*[2sinx - 1] = 0

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

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

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

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