What is the general solution for 50=25sin(pi/2 (x+1))+40 ?

The general solution for 50=25sin(pi/2 (x+1))+40 is x=(2*0.41151…)/pi+4n-1,x=4n+1-(2*0.41151…)/pi

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50=25sin(pi/2 (x+1))+40

50 = 25 sin (\frac{π}{2} (x + 1)) + 40

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

Solution steps

50 = 25 sin (\frac{π}{2} (x + 1)) + 40

Switch sides

25 sin (\frac{π}{2} (x + 1)) + 40 = 50

Move

40

to the right side

25 sin (\frac{π}{2} (x + 1)) = 10

Divide both sides by

25

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}

4 sin (2 x) = 0, - π \leq x \leq π

cos (x) = - 0.6987

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

x, y = 5 arccos (2 x)

sin (3 x) = \frac{2}{2}, 0 \leq x \leq 2 π

What is the general solution for 50=25sin(pi/2 (x+1))+40 ?
The general solution for 50=25sin(pi/2 (x+1))+40 is x=(2*0.41151…)/pi+4n-1,x=4n+1-(2*0.41151…)/pi