Solve 2-sin^2x = 2cos^2(x/2) for both x [0,2pi) and all real x.

Question

anonymous · Answer

Easier on the eye, $$2-\sin^2x = 2\cos^2(x/2)$$ I know it involves half and power reducing identities but I get stumped no matter what I do.

anonymous · Answer

$$2-\sin ^{2}x=2\cos ^{2}\frac{ x }{ 2 }$$
thus
$$2-\sin ^{2}x-1=2\cos ^{2}\frac{ x }{ 2 }-1$$
$$1-\sin ^{2}x=\cos x$$
$$\cos ^{2}x=\cos x$$
$$\cos ^{2}x-cosx=0$$
$$cosx(cosx-1)=0$$
$$cosx=0$$  or $$cosx =1$$
thus $$x=0,\frac{ \pi }{ 2 },\frac{ 3\pi }{ 2 }$$

anonymous · Answer

for real x write in genral form

anonymous · Answer

you are so kind, thank you so much!! I was making it more complicated than it needed to be.

anonymous · Answer

you re welcome

jmark · Answer

2-sin^2x = 2cos^2(x/2) 2-sin^2x - 2cos^2(x/2) =0 // subtract 2cos2(x/2) on both side // simplify and substitute y = 2cos^2(x/2) = 2- 2cos^2(x/2)-sin2x 2+3(-2)cos(1/2 x)2+(-2cos(1/2 x)2)2 = 0 y2+3y+2 = 0 // factorize // (y+1) (y+2) = 0 // split the equation // y+1 = 0 y+2 = 0 y =-1 y = -2 Substitute y= -1 in y equation -1= 2cos^2(x/2) // divide both side by -2 // ½ = cos^2(x/2) // eliminate the exponents // cos(x/2) = 1/√2 // eliminate cosine from left hand side // cos(x/2) = 1/√2 // take the inverse cosine on both side // x/2 = π/4+2πn1 for n1 b^elongs to z // multiply by 2 on both side // x = π/2+4πn1 for n2 belongs to z Learn more about Inverse Hyperbolic Sine http://goo.gl/FJQSi8