How do you solve ${sin}^{2} x = 2 cos x + 2$ $sin^2 x = 2cosx + 2$ from 0 to 2pi?

Question

3828 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-07T02:37:03

$\Rightarrow π$ $color(purple)(=> pi$

${sin}^{2} x = 2 cos x + 2$ $sin^2x = 2 cos x + 2$

$1 - {cos}^{2} x = 2 cos x + 2$ $1 - cos^2 x = 2 cos x + 2$

${cos}^{2} x + 2 cos x + 2 - 1 = 0$ $cos^2 x + 2 cos x + 2 - 1 = 0$

${cos}^{2} x + 2 cos x + 1 = 0$ $cos^2 x + 2 cos x + 1 = 0$

${(cos x = 1)}^{2} = 0$ $(cos x = 1)^2 = 0$

$cos x + 1 = 0$ $cos x + 1 = 0$

$cos x = - 1$ $cos x = -1$

$x = {cos}^{- 1} - 1 = {cos}^{- 1} cos (π) = π$ $color(purple)(x = cos ^-1 -1 = cos^-1 cos (pi) = pi$

How do you solve sin^2 x = 2cosx + 2sin2x=2cosx+2sin^2 x = 2cosx + 2 from 0 to 2pi?