Question

solve the following trigonometric equation.

Answer 1

Use:

\[\sin2x = 2sinx \cdot cosx\]

\[\implies 2sinxcosx + sinx + 2cosx + 1 = 0 \implies \sin(2x)(2 cosx + 1) + 1(2cosx + 1) = 0\]

Answer 2

Remember
\[\sin 2x = 2 \sin x\cos x\]\[\cos x=\sqrt{1-\sin^2x}\]

Answer 3

\[\implies \sin(2x)(2 cosx + 1) + 1(2cosx + 1) = 0 \implies (2cosx+1)(1 + \sin2x) = 0\]

Answer 4

wow...these replies make me nostalgic o.O i cant put a finger on it but it is as if i wrote them @_@ lol =)) jk

Answer 5

@waterineyes thanks i can do this from here :)

Answer 6

I am writing in more understandable form..

Go ahead..

Answer 7

*was

Answer 8

i do not know how to write like you
but here it is
2cos(x)+1=0
cos(x)=-1/2
then taking inverse and also the period of cos is 2npi
similarly for
1+sin(2x)
sin(2x)=-1
2x=arcsin(-1)

am i right ?

Answer 9

Yes..

Answer 10

Do you want to find general solution??

Answer 11

thanks again for your help. yes i am trying to find the general solution.
i can do this now. thank you so much,

Answer 12

Go ahead..