Since \$sin(x)=-1/2\$ is a traditional angle and also with respect come the duration I obtained \$x=pi - sin^-1(11pi/6)\$

The trouble is the I can not figure out what angle \$sin(x)=(3/2)\$.

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I have tried utilizing the Pythagorean organize but due to the fact that \$sin\$ is \$ extopp/ exthyp\$, it speak me the something is wrong because the hypotenuse can not be much shorter than the sides.

Does anyone have an idea of what I have to do?

The equation \$sin(x) = frac32\$ has actually no solution since \$-1 leq sin(x) leq 1\$ for every \$x\$.So you deserve to just solve \$sin(x) = -frac12\$ and also you"re done.

Extending to facility plane,\$\$sin^-1 frac32=fracpi2-2iln left( frac1+sqrt52 ight)\$\$

This deserve to be easily verified by using

\$\$sin (x+yi) = sin x cosh y+icos x sinh y\$\$

Solve\$\$sin(x)=frac32\$\$for \$x\$.

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Rewrite the sine function in terms of exponential function:\$\$frac32=sin(x)=frace^ix-e^-ix2i;\$\$multiply both political parties by \$2i\$:\$\$3i=e^ix-e^-ix;\$\$subtract \$3i\$ indigenous both sides:\$\$e^ix-3i-e^-ix=0;\$\$multiply both sides by \$e^ix\$:\$\$left(e^ix ight)^2-3ie^ix-e^-ixe^ix=0;\$\$rewrite \$e^-ixe^ix=1\$:\$\$left(e^ix ight)^2-3ie^ix-1=0;\$\$substitute \$u=e^ix\$:\$\$u^2-3iu-1=0;\$\$multiply both sides by \$4\$:\$\$4u^2-12iu-4=0;\$\$rewrite \$4u^2=left(2u ight)^2\$:\$\$left(2u ight)^2-12iu-4=0;\$\$add \$4+9i^2\$ come both sides:\$\$left(2u ight)^2-12iu+9i^2=4+9i^2;\$\$rewrite \$9i^2=left(3i ight)^2\$:\$\$left(2u ight)^2-12iu+left(3i ight)^2=4+9i^2;\$\$rewrite \$left(2u ight)^2-12iu+left(3i ight)^2=left(2u-3i ight)^2\$:\$\$left(2u-3i ight)^2=4+9i^2;\$\$rewrite \$4+9i^2=4-9=-5\$:\$\$left(2u-3i ight)^2=-5;\$\$take the square root of both sides:\$\$2u-3i=pm isqrt5;\$\$add \$3i\$ to both sides:\$\$2u=3ipm isqrt5;\$\$factor \$i\$ out:\$\$2u=ileft(3pmsqrt5 ight);\$\$divide both political parties by \$2\$:\$\$u=fraci2left(3pmsqrt5 ight);\$\$substitute back \$u=e^ix\$:\$\$e^ix=fraci2left(3pmsqrt5 ight);\$\$rewrite \$fraci2left(3pmsqrt5 ight)=e^logleft(3i/2pm isqrt5/2 ight)\$:\$\$e^ix=e^logleft(3i/2pm isqrt5/2 ight);\$\$since \$1=e^2ipi c_1\$ because that arbitrary creature \$c_1\$, multiply the RHS by \$e^2ipi c_1\$:\$\$e^ix=e^logleft(3i/2pm isqrt5/2 ight)e^2ipi c_1;\$\$rewrite \$e^logleft(3i/2pm isqrt5/2 ight)e^2ipi c_1=e^logleft(3i/2pm isqrt5/2 ight)+2ipi c_1\$:\$\$e^ix=e^logleft(3i/2pm isqrt5/2 ight)+2ipi c_1;\$\$eliminate exponentials:\$\$ix=logleft(frac3pmsqrt52i ight)+2ipi c_1;\$\$divide both sides by \$i\$:\$\$x=frac1ilogleft(frac3pmsqrt52i ight)+2pi c_1;\$\$rewrite \$1/i=-i\$:\$\$x=-ilogleft(frac3pmsqrt52i ight)+2pi c_1;\$\$rewrite \$logleft(frac3pmsqrt52i ight)=logleft(frac3pmsqrt52 ight)+fracipi2\$:\$\$x=-ileft(logleft(frac3pmsqrt52 ight)+fracipi2 ight)+2pi c_1;\$\$expand:\$\$x=-ilogleft(frac3pmsqrt52 ight)-fraci^2pi2+2pi c_1;\$\$rewrite \$i^2=-1\$:\$\$x=fracpi2-ilogleft(frac3pmsqrt52 ight)+2pi c_1.\$\$

From the whole collection of solutions, the principal inverse sine value is ceded by setup \$c_1=0\$ and choosing the positive debate of \$log\$:\$\$arcsinleft(frac32 ight)=oxedfracpi2-ilogleft(frac3+sqrt52 ight),\$\$where \$log\$ is the natural logarithm and also \$i\$ is the imagine unit.