Solving sin &#x2061;<!-- ⁡ --> x + sin &#x2061;<!-- ⁡ --> 2 x + sin

seupeljewj

seupeljewj

Answered question

2022-06-16

Solving sin x + sin 2 x + sin 3 x = 1 + cos x + cos 2 x

Answer & Explanation

Braedon Rivas

Braedon Rivas

Beginner2022-06-17Added 24 answers

Since
sin ( x ) + sin ( 3 x ) = 2 sin ( 2 x ) cos ( x )
and
cos ( 2 x ) = 2 cos 2 ( x ) 1 ,
one has
2 sin ( 2 x ) cos ( x ) + sin ( 2 x ) = 1 + cos ( x ) + 2 cos 2 ( x ) 1 sin ( 2 x ) ( 2 cos ( x ) + 1 ) = cos ( x ) ( 2 cos ( x ) + 1 ) ( 2 cos ( x ) + 1 ) ( sin ( 2 x ) cos ( x ) ) = 0 ( 2 cos ( x ) + 1 ) ( 2 sin ( x ) cos ( x ) cos ( x ) ) = 0 cos ( x ) ( 2 cos ( x ) + 1 ) ( 2 sin ( x ) 1 ) = 0

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?