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solve for

cos 3x=sin 2x

using complementary angles and all i got alpha=2x.

The thing is, if i got alpha 2x from sin inverse, the general formula would be

[tex] \theta= n \pi + (-1)^n \alpha [/tex]

i get something like,

so,....the problem is that i will have x in the general formuala, something i don't want.

I think my way of doing it is totally wrong, can anyone help?>

btw, i know that i can use triple and double angle formula, but i've also been taught that inversing sines both sides would also work, but i've forgotten how to do it. Thanks.

cos 3x=sin 2x

using complementary angles and all i got alpha=2x.

The thing is, if i got alpha 2x from sin inverse, the general formula would be

[tex] \theta= n \pi + (-1)^n \alpha [/tex]

i get something like,

so,....the problem is that i will have x in the general formuala, something i don't want.

I think my way of doing it is totally wrong, can anyone help?>

btw, i know that i can use triple and double angle formula, but i've also been taught that inversing sines both sides would also work, but i've forgotten how to do it. Thanks.

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