2 s in x cos x + cos x = 0 cos x ( 2 s in x + 1 ) = 0 ⟺ cos x = 0 ∨ 2 s in x + 1 = 0 cos x = 0 ∨ 2 s in x = − 1 / : 2 cos x = 0 ∨ s in x = − 2 1 x = 2 π + kπ ∨ x = − 6 π + 2 kπ ∨ x = 6 7 π + 2 kπ t o k ∈ Z
To solve the equation 2 sin x cos x + cos x = 0 , we factor it to find two conditions: cos x = 0 and sin x = − 2 1 . This gives us multiple solutions depending on the integer k .
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