Number of solution in [ 0 , 2 &#x03C0;<!-- π --> ] satisfying the equation 8 sin

Karina Trujillo 2022-06-26 Answered
Number of solution in [ 0 , 2 π ] satisfying the equation
8 sin ( x ) = 3 cos ( x ) + 1 sin ( x )
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Answers (1)

Paxton James
Answered 2022-06-27 Author has 25 answers
Multiply through by sin x cos x. We get
8 cos x ( 1 cos 2 x ) = 3 sin x + cos x .
Recall that cos 3 x = 4 cos 3 x 3 cos x. So the left-hand side becomes 8 cos x 2 ( cos 3 x + 3 cos x ), that is, 2 cos x 2 cos 3 x. Thus our equation can be rewritten as
cos 3 x = 1 2 cos x 3 2 sin x .
The right-hand side is equal to cos ( x + π / 3 ), so our equation simplifies to
cos 3 x = cos ( x + π / 3 ) .
To finish, recall that cos s = cos t if and only if s = ± t + 2 k π for some integer k.

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