Question

Solve the equation.sin x = $\frac{\sqrt{3}}{2}$

Answer 1

As we know that principal value branch of sin inverse is [-π/2,π/2]

thus while solving the equation we must remember that sin inverse cancel sin x only if x belongs to principal value branch.

$sin \: x = \frac{ \sqrt{3} }{2} \\ \\ x = {sin}^{ - 1} ( \frac{ \sqrt{3} }{2} ) \\ \\ x = {sin}^{ - 1} (sin \: ( \frac{ \pi }{3} )$
here both cancels with each other because x belongs to [-π/2,π/2]

Thus

$x = \frac{\pi}{3}$

Answer 2

Answer:

As we know that principal value branch of sin inverse is [-π/2,π/2]

thus while solving the equation we must remember that sin inverse cancel sin x only if x belongs to principal value branch.

\begin{gathered}sin \: x = \frac{ \sqrt{3} }{2} \\ \\ x = {sin}^{ - 1} ( \frac{ \sqrt{3} }{2} ) \\ \\ x = {sin}^{ - 1} (sin \: ( \frac{ \pi }{3} ) \\ \\\end{gathered}

sinx=

2

3

x=sin

−1

(

2

3

)

x=sin

−1

(sin(

3

π

)

here both cancels with each other because x belongs to [-π/2,π/2]

Thus

\begin{gathered}x = \frac{\pi}{3} \\ \\\end{gathered}

x=

3

π