question

question  x = ?

sinx +cosx = √6/2

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\(sinx+cosx=\frac{1}{2}\sqrt{2}\\ sinx+\sqrt{1-sin^2x}=\frac{1}{2}\sqrt{2}\\ sinx=u\\ u+\sqrt{1-u^2}=\frac{1}{2}\sqrt{2}\\ \sqrt{1-u^2}=\frac{1}{2}\sqrt{2}-u\)        \(sinx-\sqrt{1-sin^2x}=\frac{1}{2}\sqrt{2}\)   (also pursue!)

\(1-u^2=\frac{1}{2}-\sqrt{2}\cdot u+u^2\\ 2u^2-\sqrt{2}\cdot u-\frac{1}{2}=0\\ u^2-\frac{1}{2}\sqrt{2}\cdot u-\frac{1}{4}=0 \)

\(u=\frac{1}{4}\sqrt{2}\pm \sqrt{\frac{1}{8}+\frac{1}{4}}\\ u=\frac{1}{4}\sqrt{2}\pm \sqrt{\frac{3}{8}}\\\)

\(u_1=sinx_1=0.96592\\ u_2=sinx_2=-0.25882\)

\(x_1=105°\\ x_2=-15°\)

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