What is the general solution for sin(pi/2 (x+1))= 2/5 ?

The general solution for sin(pi/2 (x+1))= 2/5 is x=(2*0.41151…)/pi+4n-1,x=4n+1-(2*0.41151…)/pi

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sin(pi/2 (x+1))= 2/5

sin (\frac{π}{2} (x + 1)) = \frac{2}{5}

x = \frac{2 \cdot 0.41151 \dots}{π} + 4 n - 1, x = 4 n + 1 - \frac{2 \cdot 0.41151 \dots}{π}

Solution steps

sin (\frac{π}{2} (x + 1)) = \frac{2}{5}

Apply trig inverse properties

\frac{π}{2} (x + 1) = arcsin (\frac{2}{5}) + 2 πn, \frac{π}{2} (x + 1) = π - arcsin (\frac{2}{5}) + 2 πn

Solve

\frac{π}{2} (x + 1) = arcsin (\frac{2}{5}) + 2 πn : x = \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n - 1

Solve

\frac{π}{2} (x + 1) = π - arcsin (\frac{2}{5}) + 2 πn : x = 4 n + 1 - \frac{2 arcsin ( \frac{2}{5} )}{π}

x = \frac{2 arcsin ( \frac{2}{5} )}{π} + 4 n - 1, x = 4 n + 1 - \frac{2 arcsin ( \frac{2}{5} )}{π}

Show solutions in decimal form

x = \frac{2 \cdot 0.41151 \dots}{π} + 4 n - 1, x = 4 n + 1 - \frac{2 \cdot 0.41151 \dots}{π}

sin (\cdot + \frac{π}{2}) = cos (x)

sin (θ) = (0.000069) 2 (500 \cdot 1 0^{- 9})

(2 sin^{2} (x) - 1) (tan (x) - 1) = 0

(csc (x) + 2) (csc (x) - 2) = 0, 0 \leq x \leq 2 π

cos (x) = 55

What is the general solution for sin(pi/2 (x+1))= 2/5 ?
The general solution for sin(pi/2 (x+1))= 2/5 is x=(2*0.41151…)/pi+4n-1,x=4n+1-(2*0.41151…)/pi