Solve:

Asked by Sneha George | 26th Jul, 2011, 11:37: PM

Expert Answer:

a. sin3x+cos2x=0
sin(2x+x)+cos2x=0
sin2x.cosx+cos2x.sinx+cos2x=0
2sinx.cos^2(x)+cos^2(x).sinx-sin^3(x)-…
3cos^2(x).sinx-sin^3(x)-1=0
sinx[3cos^2(x)-sin^2(x)]-1=0 let cos^2(x)=1-sin^2(x)
sinx[3-4sin^2(x)]-1=0
3sinx-4sin^3(x)-1=0 many solutions might exist.
one solution is sinx=-1
x=3pi/4+2piK where K is an integer.

b. Cos3x = sin2x
 Cos3x = cos(r/2 - 2x)
 3x = r/2 - 2x + k2r ( k of Z )
 3x = -r/2 + 2x + k2r
 x = r/10 + k2r /5
 x = -r/2 + k2r

e. replace cos^2 with 1-sin^2 x
2(1-sin^x) + 3sinx = 0
-2sin^2x + 3sinx + 2 = 0

2sin^2x -3sinx - 2 = 0
(2sinx + 1)(sinx - 2) = 0
sinx = -1/2 or sinx =2 ---but sinx cannot = 2 so just work with -1/2

sinx = -1/2 quadrants III and IV
x = 210 deg (or 7pi/6) + 2npi in QIII
x = 330 deg (or 11pi/6) + 2npi in QIV

f. Tanx=sinx
sinx/cosx=sinx
sinx=sinx*cosx
0=sinx*cosx-sinx
0=sinx(cosx-1)
sinx=0     cosx=1
x1=k*pi    x2=2*k*pi ; k € IR
x1 covers both so x=k*pi.

 

 

Answered by  | 27th Jul, 2011, 02:55: PM

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