4 x 1 − x 2 − 2 = 0
For x ∈ [ − 1 , 1 ] , the roots of the equation above can be expressed in the form cos ( n 1 π ) and sin ( n 1 π ) , where n is a positive integer.
Find the value of n .
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I suppose we could also have that sin ( 2 θ ) = 2 1 ⟹ 2 θ = 4 3 π ⟹ θ = 8 3 π ,
which still meets the condition that 0 ≤ θ ≤ 2 π . This would yield an alternate answer of 8 − 3 = 5 . The pairs of roots are identical since sin ( 8 π ) = cos ( 8 3 π ) and cos ( 8 π ) = sin ( 8 3 π ) , but the solution to this question as presently worded is not unique.
note that x must be positive. now solve 4 x 1 − x 2 = 2 1 6 x 2 − 1 6 x 4 = 2 8 x 4 − 8 x 2 + 1 = 0 x 2 = 4 2 ± 2 x = 2 2 ± 2 negative neglected. since in the first quadrant sin (y)<cos (y), c o s ( y ) = 2 2 + 2 , s i n ( y ) = 2 2 − 2 c o s ( 2 y ) = c o s 2 ( y ) − s i n 2 ( y ) = 4 2 + 2 − 4 2 − 2 = 2 2 2 y = 4 π ⟹ y = 8 1 π 8 − 1 = 7 not that other solutions to 2y neglected as conditions for first quadrant.
x ∈ [ − 1 , 1 ] h i n t s u s t o s u b s t i t u t e x = c o s θ . 1 − x 2 = 1 − c o s 2 θ = ∣ s i n θ ∣ s i n c e θ ∈ 1 s t q u a d r a n t ∣ s i n θ ∣ = s i n θ ∴ 4 c o s θ s i n θ = 2 s i n 2 θ = 2 1 s i n 2 θ = s i n 4 π θ = 8 π x = c o s 8 π a = 1 , b = 8 b − a = 7 .
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Let x = sin θ as suggested, then:
4 x 1 − x 2 − 2 4 x 1 − x 2 4 sin θ 1 − sin 2 θ 4 sin θ cos θ 2 sin ( 2 θ ) sin ( 2 θ ) ⇒ 2 θ θ = 0 = 2 = 2 = 2 = 2 = 2 1 = 4 π = 8 π for 0 ≤ θ ≤ 2 π
⇒ b − a = 8 − 1 = 7