What is the general solution for 5sin(y)cos(y)=2 ?

The general solution for 5sin(y)cos(y)=2 is y=(0.92729…)/2+pin,y= pi/2-(0.92729…)/2+pin

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5sin(y)cos(y)=2

5 sin (y) cos (y) = 2

y = \frac{0.92729 \dots}{2} + πn, y = \frac{π}{2} - \frac{0.92729 \dots}{2} + πn

Solution steps

5 sin (y) cos (y) = 2

Rewrite using trig identities

5 \cdot \frac{sin ( 2 y )}{2} = 2

Refine

5 \cdot \frac{sin ( 2 y )}{2} : \frac{5 sin ( 2 y )}{2}

\frac{5 sin ( 2 y )}{2} = 2

Multiply both sides by

2

5 sin (2 y) = 4

Divide both sides by

5

sin (2 y) = \frac{4}{5}

Apply trig inverse properties

2 y = arcsin (\frac{4}{5}) + 2 πn, 2 y = π - arcsin (\frac{4}{5}) + 2 πn

Solve

2 y = arcsin (\frac{4}{5}) + 2 πn : y = \frac{arcsin ( \frac{4}{5} )}{2} + πn

Solve

2 y = π - arcsin (\frac{4}{5}) + 2 πn : y = \frac{π}{2} - \frac{arcsin ( \frac{4}{5} )}{2} + πn

y = \frac{arcsin ( \frac{4}{5} )}{2} + πn, y = \frac{π}{2} - \frac{arcsin ( \frac{4}{5} )}{2} + πn

Show solutions in decimal form

y = \frac{0.92729 \dots}{2} + πn, y = \frac{π}{2} - \frac{0.92729 \dots}{2} + πn

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What is the general solution for 5sin(y)cos(y)=2 ?
The general solution for 5sin(y)cos(y)=2 is y=(0.92729…)/2+pin,y= pi/2-(0.92729…)/2+pin