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.