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Solve
$$ \sin x + \sqrt{ \sin x + \sin 2x  - \cos x } = \cos x   $$

Move the $\sin x$ to the other side, and square. Squaring will introduce some extra solutions, but worry about that later.

$$ \sin x + \sin {2x} - \cos x = {(\cos x - \sin x)}^2 $$

Not let $x = y + \pi/4$. This gives $\sin x = {\frac 1 {\sqrt{2}}} ( \sin y + \cos y )$ and 
$\cos x = {\frac 1 {\sqrt{2}}} ( \cos y - \sin y )$ and $\sin 2x = \cos 2y$. Substituting gives

$$ \sqrt 2 \sin y + \cos 2y = 2 \sin^2 y $$

Using double angle formula $\cos 2y = 1 - 2 \sin^2 y$

$$ 4 \sin^2 y - \sqrt 2 \sin y - 1 = 0 $$

Use quadratic formula to give

$$ \sin y = \frac {\sqrt 2 \pm \sqrt{2 + 4 \times 4}}{8} $$
 
which simplifies to

$$ \sin y = \frac {\sqrt 2 ( 1 \pm 3 ) } {8} $$

and finally

$$ \sin y = \frac {\sqrt 2} 2 \; \text{or} \;  - \frac {\sqrt 2}{4}$$
$$ \cos y = \pm \frac {\sqrt 2} 2 \; \text{or} \; \pm \sqrt{\frac {7}{8}}$$

The case $\sin y = 1/sqrt 2$ and $\cos y = - 1/\sqrt 2$ gives $\sin x = 0$ and $\cos x = -1$ which is an extra solution introduced in the squaring.

I think the other solution based on $\sin y = -\frac 1 {2 \sqrt 2}$ and $\cos y = \pm \frac {\sqrt 7} {2 \sqrt 2}$ is real.

Sun, 16 Feb 2025 12:11 GMT