need help finding the solutions that are in the interval [0,2Pi) for cos2x+3cosx+2=0?

Answer 1

COS2X=2COS^2X -1 ,THEN WE NOTE ;COSX=Y AND WE HAVE
2Y^2-1+3Y+2=0
2Y^2+3Y+1=0 => Y=-1 OR Y=-1/2 THEN COSX=-1 => X=PI
COSX=-1/2 => X=2PI/3. ok????

Answer 2

ok its kinda complicated yet gets extremely merely on the suitable sinxcos2x+cosxsin2x=0 we going to be employing the sin^2x +cos^2=a million alot cos2x=sin^2 - cos^2 and sin2x=2sinxcosx sinx(sin^2x - cos^2x) + cosx(2sinxcosx)=0 sinx(sin^2x-a million+sin^2x)+2sinxcos^2x=0 2sin^3x-sinx+2sinx(a million-sin^2x)= 2sin^3x-sinx+2sinx-2sin^3x=0 sinx=0 notice how the sines cancel one yet yet another out and from user-friendly understanding 0,a hundred 80 and 360 furnish 0 so the respond is 0,pi,2pi sumtimes u merely get fortunate fixing them yet very practically consistently the extra ideal subject concerns u artwork the less complicated it gets to come back to a determination comparable ones

Answer 3

Begin with cos 2x = cos(x+x) = (cos x)^2 - (sin x)^2
Then use sin^2 + cos^2 = 1 to change the (sin x)^2 into a cos function.

From then on, it is a quadratic where the "variable" is a trig function