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( 1 + sin x 1 ) cos x + ( 1 + cos x 1 ) sin x cos x + cot x + sin x + tan x sin x + cos x − 2 2 sin ( x + 4 π ) − 2 = 2 + 2 = 2 + 2 = 2 − tan x − cot x = 2 − ( tan x + cot x )
Note that − 2 ≤ 2 sin ( x + 4 π ) ≤ 2 then the left-hand side is − 2 2 ≤ L H S ≤ 0 . For 0 ≤ x < 2 π L H S = 0 , when x = 4 π .
Considering tan x + cot x > 0 , by AM-GM inequality tan x + cot x ≥ 2 tan x cot x = 2 . Then the right-hand side, R H S ≤ 0 . For 0 ≤ x < 2 π , R H S = 0 , when x = 4 π and x = 4 5 π . Therefore, L H S = R H S , when x = 4 π and a value of sin x = sin 4 π = 2 1 . Since the R H S = − 2 2 , when x ≈ 0 . 2 1 4 , 1 . 3 5 7 , 3 . 3 5 5 , 4 . 4 9 9 , there are two other points when L H S = R H S . Therefore there are 3 possible values of sin x .