How do you solve $cos 2x- sin^2 (x/2) + 3/4 = 0$ ?

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Answer 1 · 2015-08-10T15:06:12

$Cosx=1/2$ and $cosx=-3/4$

Step 1:
$cos2x-Sin^2(x/2)+3/4=0$
Use $cos2x=cos^2x-sin^2x$

Step 2:
$cos^2x-sin^2x-sin^2(x/2)+3/4=0$
Use $sin^2x+cos^2x=1$

Step3:
$2cos^2x-1-sin^2(x/2)+3/4=0$
Use $cosx=1-2sin^2(x/2)$ (Double angle formula).

Step 4:
$2cos^2x-1-1/2+1/2cosx+3/4=0$
$2cos^2x+2cosx-3=0$

Multiply by 4 to get

$8cos^x+2cosx-3=0$

Step 5: Solve the quadratic equation to get

$(2cos-1)(4cosx+3)=0$
$cosx=1/2$ and $cosx=-3/4$

How do you solve cos 2x- sin^2 (x/2) + 3/4 = 0cos 2x- sin^2 (x/2) + 3/4 = 0?