What is the general solution for solvefor α,sin^2(3α)=0.56 ?

The general solution for solvefor α,sin^2(3α)=0.56 is

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solvefor α,sin^2(3α)=0.56

solve for

α, sin^{2} (3 α) = 0.56

α = \frac{0.84554 \dots}{3} + \frac{2 πn}{3}, α = \frac{π}{3} - \frac{0.84554 \dots}{3} + \frac{2 πn}{3}, α = - \frac{0.84554 \dots}{3} + \frac{2 πn}{3}, α = \frac{π}{3} + \frac{0.84554 \dots}{3} + \frac{2 πn}{3}

Solution steps

sin^{2} (3 α) = 0.56

Solve by substitution

sin (3 α) = 0.56, sin (3 α) = - 0.56

sin (3 α) = 0.56 : α = \frac{arcsin ( 0.56 )}{3} + \frac{2 πn}{3}, α = \frac{π}{3} - \frac{arcsin ( 0.56 )}{3} + \frac{2 πn}{3}

sin (3 α) = - 0.56 : α = - \frac{arcsin ( \frac{14}{5} )}{3} + \frac{2 πn}{3}, α = \frac{π}{3} + \frac{arcsin ( 0.56 )}{3} + \frac{2 πn}{3}

Combine all the solutions

α = \frac{arcsin ( 0.56 )}{3} + \frac{2 πn}{3}, α = \frac{π}{3} - \frac{arcsin ( 0.56 )}{3} + \frac{2 πn}{3}, α = - \frac{arcsin ( \frac{14}{5} )}{3} + \frac{2 πn}{3}, α = \frac{π}{3} + \frac{arcsin ( 0.56 )}{3} + \frac{2 πn}{3}

Show solutions in decimal form

α = \frac{0.84554 \dots}{3} + \frac{2 πn}{3}, α = \frac{π}{3} - \frac{0.84554 \dots}{3} + \frac{2 πn}{3}, α = - \frac{0.84554 \dots}{3} + \frac{2 πn}{3}, α = \frac{π}{3} + \frac{0.84554 \dots}{3} + \frac{2 πn}{3}

(sin (x) - cos (x))^{2} = 1^{2}

tan (\frac{π}{2} - x) + 2 tan (\frac{π}{2} + x) = 1

\frac{11.0994}{cos ( x )} = \frac{3.5355}{sin ( x )}

(\frac{sec ( x )}{cos ( x )})^{2} = tan^{2} (x)

sin (\frac{x}{2}) - cos (\frac{x}{2}) = 0

What is the general solution for solvefor α,sin^2(3α)=0.56 ?
The general solution for solvefor α,sin^2(3α)=0.56 is