a trigonometry problem

Geometry Level 4

cos ⁡ ( 5 x ) + cos ⁡ ( x ) + 2 cos ⁡ ( 3 x ) sin ⁡ ( 2 x ) = 12 cos ⁡ ( 3 x ) \large \cos { (5x) } +\cos { (x) } +2\cos { (3x) } \sin { (2x) } =12\cos { (3x) }

If the solution of the equation above for 0 < x < π 2 0<x<\dfrac { \pi }{ 2 } can be written as a b π \dfrac { a }{ b } \pi , where a a and b b are coprime positive integers, find a + b a+b .

Source : adapted from 1971's Lomonosov Moscow State University physics course admission exam.


The answer is 7.

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João Vitor Cordeiro de Brito by João Vitor Cordeiro de B… , 27

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1 solution

Note that c o s 5 x + c o s x = 2 cos ⁡ 3 x cos ⁡ 2 x cos{5x}+cos{x}=2\cos{3x}\cos{2x} Hence, the given equation becomes 2 cos ⁡ 3 x ( sin ⁡ 2 x + cos ⁡ 2 x − 6 ) = 0 2\cos{3x}(\sin{2x}+\cos{2x}-6)=0 As sin ⁡ 2 x + cos ⁡ 2 x ≤ 2 < 6 \sin{2x}+\cos{2x}\leq\sqrt{2}<6 We have, cos ⁡ 3 x = 0 \cos{3x}=0 . Now, 0 < x < π 2 ⇒ 0 < 3 x < 3 π 2 ⇒ 3 x = π 2 ⇒ x = π 6 0<x<\frac{\pi}{2}\Rightarrow0<3x<\frac{3π}{2}\Rightarrow3x=\frac{\pi}{2}\Rightarrow x=\frac{\pi}{6} So, a = 1 , b = 6 ⇒ a + b = 7 \boxed{a=1, b=6\Rightarrow a+b=7}

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