Sin 45+tan 75+sin X^3 = 4.5

Were you supposed to simply this and get an exact answer?

sin45=1/(sqrt2)

tan75=tan(60+15) you can expand this 

To get the ratios for 15 degrees you can use half angle forumulas which are included on this page

http://www.sosmath.com/trig/douangl/douangl.html

If you are really keen there is a long version here

http://www.cut-the-knot.org/pythagoras/cos15.shtml

that should help you along the way!

Want more help - just ask!

Using the calculator here:

$${\mathtt{sinX}} = {\left({\mathtt{4.5}}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{45}}^\circ\right)}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{75}}^\circ\right)}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} \Rightarrow {\mathtt{sinX}} = {\mathtt{0.393\: \!310\: \!438\: \!163\: \!336\: \!6}}$$

$${\mathtt{X}} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\mathtt{0.393\: \!310\: \!438\: \!163\: \!336\: \!6}}\right)} \Rightarrow {\mathtt{X}} = {\mathtt{23.160\: \!642\: \!003\: \!775^{\circ}}}$$

X ≈ 23.16°

Could also have X = 180°-23.16° as sin is positive in the second quadrant.