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$sin 2 x - 2 cos x + 0.25 = 0$ $sin2x-2cosx+0.25=0$

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Answer 1 · 2018-01-24T12:10:22

$sin (2 x) = 2 sin x cos x$ $sin(2x)=2sinx cosx$ then

$sin 2 x - 2 cos x + 0.25 = 2 sin x cos x - 2 cos x + \frac{1}{4} = 0$ $sin2x-2cosx+0.25=2sinx cosx-2cosx+1/4=0$

then

$cos x = \frac{1}{8 (sin x - 1)}$ $cos x = 1/(8(sinx-1))$ but

${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x = 1$ so

${sin}^{2} x + \frac{1}{64 {(sin x - 1)}^{2}} = 1$ $sin^2x+1/(64(sinx-1)^2)=1$ and now solving for $sin x$ $sin x$ we have the real solutions as

$sinx = {(-0.998041),(0.794264):}$

The next steps are left to the reader.