What is the general solution for 6sin^2(4x)-4=0 ?

The general solution for 6sin^2(4x)-4=0 is

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6sin^2(4x)-4=0

6 sin^{2} (4 x) - 4 = 0

x = \frac{0.95531 \dots}{4} + \frac{πn}{2}, x = \frac{π}{4} - \frac{0.95531 \dots}{4} + \frac{πn}{2}, x = - \frac{0.95531 \dots}{4} + \frac{πn}{2}, x = \frac{π}{4} + \frac{0.95531 \dots}{4} + \frac{πn}{2}

Solution steps

6 sin^{2} (4 x) - 4 = 0

Solve by substitution

sin (4 x) = \frac{2}{3}, sin (4 x) = - \frac{2}{3}

sin (4 x) = \frac{2}{3} : x = \frac{arcsin ( \frac{2}{3} )}{4} + \frac{πn}{2}, x = \frac{π}{4} - \frac{arcsin ( \frac{2}{3} )}{4} + \frac{πn}{2}

sin (4 x) = - \frac{2}{3} : x = - \frac{arcsin ( \frac{2}{3} )}{4} + \frac{πn}{2}, x = \frac{π}{4} + \frac{arcsin ( \frac{2}{3} )}{4} + \frac{πn}{2}

Combine all the solutions

x = \frac{arcsin ( \frac{2}{3} )}{4} + \frac{πn}{2}, x = \frac{π}{4} - \frac{arcsin ( \frac{2}{3} )}{4} + \frac{πn}{2}, x = - \frac{arcsin ( \frac{2}{3} )}{4} + \frac{πn}{2}, x = \frac{π}{4} + \frac{arcsin ( \frac{2}{3} )}{4} + \frac{πn}{2}

Show solutions in decimal form

x = \frac{0.95531 \dots}{4} + \frac{πn}{2}, x = \frac{π}{4} - \frac{0.95531 \dots}{4} + \frac{πn}{2}, x = - \frac{0.95531 \dots}{4} + \frac{πn}{2}, x = \frac{π}{4} + \frac{0.95531 \dots}{4} + \frac{πn}{2}

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What is the general solution for 6sin^2(4x)-4=0 ?
The general solution for 6sin^2(4x)-4=0 is