A geometry problem by A Former Brilliant Member

Geometry Level 3

Find the number of integral values of k k for which the equation 7 cos ⁡ x + 5 sin ⁡ x = 2 k + 1 7 \cos x + 5 \sin x = 2k +1 has a solution.

6 7 1 No solution 2 8

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A Former Brilliant Member by A Former Brilliant Member

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1 solution

Sabhrant Sachan
Sep 30, 2016

Multiply and divide by 7 2 + 5 2 = 74 Let t = 74 ≈ 8.602 t ( 7 t cos ⁡ x + 5 t sin ⁡ x ) = 2 k + 1 Let α = sin ⁡ − 1 ( 7 t ) t sin ⁡ ( x + α ) = 2 k + 1 x + α = sin ⁡ − 1 ( 2 k + 1 t ) ⟹ − 1 ≤ 2 k + 1 t ≤ 1 ⟹ − 1 2 ( t + 1 ) ≤ k ≤ 1 2 ( t − 1 ) k ∈ [ − 4.801 , 3.801 ] k = − 4 , − 3 , − 2 , − 1 , 0 , 1 , 2 , 3 8 possible values of k \text{Multiply and divide by } \sqrt{7^2+5^2} = \sqrt{74} \\ \text{Let } \color{#3D99F6}{t}=\sqrt{74} \approx 8.602 \\ \color{#3D99F6}{t}\left( \dfrac{7}{\color{#3D99F6}{t}}\cos{x}+\dfrac{5}{\color{#3D99F6}{t}}\sin{x} \right) = 2k+1 \\ \text{Let } \alpha = \sin^{-1}{\left( \dfrac{7}{\color{#3D99F6}{t}} \right) } \\ \color{#3D99F6}{t}\sin(x+\alpha)=2k+1 \\ x+\alpha = \sin^{-1}{\left(\dfrac{2k+1}{\color{#3D99F6}{t}} \right)} \\ \implies -1 \le \dfrac{2k+1}{\color{#3D99F6}{t}} \le 1 \implies -\dfrac{1}{2}\left( \color{#3D99F6}{t}+1 \right)\le k \le \dfrac{1}{2}\left( \color{#3D99F6}{t}-1\right) \\ k \in \left[-4.801,3.801\right] \\ k=-4,-3,-2,-1,0,1,2,3 \\ \boxed{8}\text{ possible values of k }

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Another easier way is to use Cauchy Schwarz Inequality.

A Former Brilliant Member - 4 years, 8 months ago

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Please Post your Solution :) .

Sabhrant Sachan - 4 years, 8 months ago

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7 cos ⁡ ( x ) + 5 sin ⁡ ( x ) ≤ 7 2 + 5 2 cos ⁡ 2 ( x ) + sin ⁡ 2 ( x ) = 74 7\cos(x) + 5\sin(x) \le \sqrt{7^2+5^2}\sqrt{\cos^2(x)+\sin^2(x)} = \sqrt{74}

Multiply the equation by − 1 -1 and you get:

− 74 ≤ − 7 cos ⁡ ( x ) − 5 sin ⁡ ( x ) = 7 cos ⁡ ( x + π ) + 5 sin ⁡ ( x + π ) -\sqrt{74} \le -7\cos(x) - 5\sin(x) = 7\cos(x+\pi) + 5\sin(x+\pi)

Hence, for all x x , we have − 74 ≤ 7 cos ⁡ ( x ) + 5 sin ⁡ ( x ) ≤ 74 -\sqrt{74} \le 7\cos(x) + 5\sin(x) \le \sqrt{74}

Ariel Gershon - 4 years, 8 months ago

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@Ariel Gershon – Exactly!(+1)

A Former Brilliant Member - 4 years, 8 months ago

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