What is the general solution for solvefor x,arcsin(3x-4y)= pi/2 ?

The general solution for solvefor x,arcsin(3x-4y)= pi/2 is x= 1/3+(4y)/3

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solvefor x,arcsin(3x-4y)= pi/2

solve for

x, arcsin (3 x - 4 y) = \frac{π}{2}

x = \frac{1}{3} + \frac{4 y}{3}

Solution steps

arcsin (3 x - 4 y) = \frac{π}{2}

Apply trig inverse properties

arcsin (3 x - 4 y) = \frac{π}{2}

arcsin (x) = a \Rightarrow x = sin (a)

3 x - 4 y = sin (\frac{π}{2})

sin (\frac{π}{2}) = 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

sin (\frac{π}{2})

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 1

= 1

3 x - 4 y = 1

3 x - 4 y = 1

Solve

3 x - 4 y = 1 : x = \frac{1}{3} + \frac{4 y}{3}

3 x - 4 y = 1

Move

4 y

to the right side

3 x - 4 y = 1

Add

4 y

to both sides

3 x - 4 y + 4 y = 1 + 4 y

Simplify

3 x = 1 + 4 y

3 x = 1 + 4 y

Divide both sides by

3

3 x = 1 + 4 y

Divide both sides by

3

\frac{3 x}{3} = \frac{1}{3} + \frac{4 y}{3}

Simplify

x = \frac{1}{3} + \frac{4 y}{3}

x = \frac{1}{3} + \frac{4 y}{3}

x = \frac{1}{3} + \frac{4 y}{3}

sin (x) a = \frac{π}{3}

sin (x - \frac{π}{2}) = cos (x)

cos (θ) = \frac{6}{14 \cdot 9}

sin (θ + \frac{π}{4}) = \frac{1}{2}

(cot (x) - 1) (sin (x) - 1) = 0

What is the general solution for solvefor x,arcsin(3x-4y)= pi/2 ?
The general solution for solvefor x,arcsin(3x-4y)= pi/2 is x= 1/3+(4y)/3