trig equation qn (2 Viewers)

[Masaken]

Yeah I got +- pi/2 +2npi, but theres also the other two solutions for pi/6+2npi tho...

so when there's cos2x + |sinx| = 0, one of the cases is sinx <0, where x is in the domain of (pi, 2pi) (for sinx is equal than or > 0, x is in the domain [0, pi].

you get cos2x - sinx = 0, then simplify that to a quadratic which factors to (sinx+1)(2sinx-1) = 0. bear in mind since sin x < 0 and considering the given restrictions sin x has to be between [-1,0]. so when you get sinx=-1 and sinx=1/2, you can only take -1 as 1/2 lies outside that region, and thus pi/6 isn't a solution, same applies for sinx > 0 (pretend there's a line under the >) except you would get sinx=1 and sinx=-1/2

 

[Average Boreduser]

so when there's cos2x + |sinx| = 0, one of the cases is sinx <0, where x is in the domain of (pi, 2pi) (for sinx is equal than or > 0, x is in the domain [0, pi].

you get cos2x - sinx = 0, then simplify that to a quadratic which factors to (sinx+1)(2sinx-1) = 0. bear in mind since sin x < 0 and considering the given restrictions sin x has to be between [-1,0] because of the unit circle. so when you get sinx=-1 and sinx=1/2, you can only take -1 as 1/2 lies outside that region, and thus pi/6 isn't a solution, same applies for sinx > 0 (pretend there's a line under the >) except you would get sinx=1 and sinx=-1/2

I just died of death from math overload. I'm gonna have sit down.

 

[Masaken]

I just died of death from math overload. I'm gonna have sit down.

i'm not very good at explaining, i apologise

 

[5uckerberg]

Yeah I got +- pi/2 +2npi, but theres also the other two solutions for pi/6+2npi tho...

The problem is that is actually 1 which is