missing solutions for trig equation?? (1 Viewer)

[get_back23]

say for 2 sin squared x = -sinx
i divide by sinx
becomes 2sinx = -1
but then i would miss a solution comparing to if i did 2sinsquared x + sinx =0 then factorise

... and how do i know when to use this method to not miss a solution?, or should i just ignore it

Thanks so much

 

[nightweaver066]

say for 2 sin squared x = -sinx
i divide by sinx
becomes 2sinx = -1
but then i would miss a solution comparing to if i did 2sinsquared x + sinx =0 then factorise

... and how do i know when to use this method to not miss a solution?, or should i just ignore it

Thanks so much

Never do it.

If you divide both sides by sinx, you're assuming that sinx cannot equal to 0 (you cannot divide by 0) so you end up losing a solution (or multiple solutions depending on the domain).

 

[Aesytic]

it's probably best to avoid dividing by an unknown when solving an equation, but if you wanted to, i suppose you could test the solution sinx=0, and if it didn't work, then you could divide the whole equation by sinx

 

[Shadowdude]

Never do it.

If you divide both sides by sinx, you're assuming that sinx cannot equal to 0 (you cannot divide by 0) so you end up losing a solution (or multiple solutions depending on the domain).

^ That.

just factorise 'em out

 

[asianese]

If you divide by sinx it's like dividing out x from but obviously as well!

 

[get_back23]

oh i c

but then doesnt that mean its harder to solve those trig equations without powers, because what can i do now? factorise if its easy, but otherwise its t/2 and auxillary?

 

[theind1996]

oh i c

but then doesnt that mean its harder to solve those trig equations without powers, because what can i do now? factorise if its easy, but otherwise its t/2 and auxillary?

Can you please give me a "harder trig equation without powers"?

Because this one can be easily factorised and treated like a Quadratic..