How do you solve for x? 0<x<2 pi .... sinxcos2 + cosxsin2 = -0.5?

Question

My teacher gave us this trig assignment and its like an ISU, because he will never teach it to us...so please explain HOW you get this answer...so i can do the rest...THANKS!

Answer 1

using sin (A+B) = sin A cos B + cos A sin B we recoginze

LHS = sin(x+2) = -.5 = sin (-pi/6) = sin 7pi/6 or sin 11pi/6

so x = 7pi/6-2 or 11pi/6- 2

Answer 2

Before I can answer, please clarify: is it really
sin(x) cos(2) + cos(x) sin(2) = -0.5
or
sin(x) cos(2x) + cos(x) sin(2x) = -0.5
or
sin(x) cos^2(x) + cos(x) sin^2(x) = -0.5
?

A typical trig problem would never use "cos2" or "sin2".

In any case, what you would normally do is use trigonometric identities to rewrite the equation so that it's easier to solve, usually by factoring out a sin(x) or cos(x) or similar expression. You end up with equations of the form sin(x) = ? or cos(x) = ?, for which you use your knowledge of sin and cos at "basic" angles like pi/6, pi/4, pi/2, etc. as well as the symmetries of sin and cos, such as sin(pi-x) = sin(x), to find the possible values of x.