Question

Solve for $\alpha$:

$0.42 \alpha = \left( \sin^2 \alpha - 0.07 \sin 2\alpha \right) \left( \frac{2 \sin^2 \alpha - 0.14 \sin 2\alpha}{2 \sin 2\alpha - 0.14 \cos 2\alpha} \right)^2$

Answer 1

$\alpha = 0$

Answer 2

Let $k = 0.07$, then the equation becomes:

$6k \alpha = \left( \sin^2 \alpha - k \sin 2\alpha \right) \left( \frac{2 \sin^2 \alpha - 2k \sin 2\alpha}{2 \sin 2\alpha - 2k \cos 2\alpha} \right)^2$

Let $A = \sin^2 \alpha - k \sin 2\alpha$. Then $A' = \frac{dA}{d\alpha} = \sin 2\alpha - 2k \cos 2\alpha$.

The equation can be written as:

$6k \alpha = A \left( \frac{2A}{A'} \right)^2 = \frac{4A^3}{(A')^2}$

$6k \alpha (A')^2 = 4A^3$

Taking the square root:

$\sqrt{6k \alpha} A' = \pm 2A^{3/2}$

$\sqrt{6k \alpha} \frac{dA}{d\alpha} = \pm 2A^{3/2}$

Separating variables:

$\frac{dA}{A^{3/2}} = \pm \frac{2}{\sqrt{6k \alpha}} d\alpha$

Integrating both sides:

$\int A^{-3/2} dA = \pm \frac{2}{\sqrt{6k}} \int \alpha^{-1/2} d\alpha$

$-2A^{-1/2} = \pm \frac{4}{\sqrt{6k}} \alpha^{1/2} + C$

$-\frac{2}{\sqrt{A}} = \pm \frac{4}{\sqrt{6k}} \sqrt{\alpha} + C$

If $\alpha = 0$, then:

$0.42 \times 0 = (\sin^2 0 - 0.07 \sin 0) \left( \frac{2 \sin^2 0 - 0.14 \sin 0}{2 \sin 0 - 0.14 \cos 0} \right)^2$

$0 = (0 - 0) \left( \frac{0 - 0}{0 - 0.14} \right)^2 = 0$

Therefore, $\alpha = 0$ is a solution.