Tricky Trigonometry

Geometry Level pending

x x is a real number. Let S S be the sum of all the possible distinct values of sin ⁡ x \sin x that satisfy the following equation:

7 − 2 cos ⁡ 2 x − 11 sin ⁡ x = 0 7 - 2\cos^2x - 11\sin x = 0

S S can be written as a b \frac{a}{b} , where a a and b b are coprime positive integers. What is the value of a + b a + b ?


The answer is 3.

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Calvin Lin by Calvin Lin

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1 solution

Calvin Lin Staff
May 13, 2014

Substituting in cos ⁡ 2 x = 1 − sin ⁡ 2 x \cos^2x = 1 - \sin^2x , we have

0 = 7 − 2 cos ⁡ 2 x − 11 sin ⁡ x = 7 − 2 ( 1 − sin ⁡ 2 x ) − 11 sin ⁡ x = 5 + 2 sin ⁡ 2 x − 11 sin ⁡ x = ( 2 sin ⁡ x − 1 ) ( sin ⁡ x − 5 ) \begin{aligned} 0 &= 7 - 2\cos^2x - 11\sin x \\ &= 7 - 2(1 - \sin^2x) - 11 \sin x \\ &= 5 + 2 \sin^2 x - 11\sin x \\ &= (2\sin x - 1)(\sin x - 5) \\ \end{aligned}

Thus sin ⁡ x = 1 2 \sin x = \frac{1}{2} and sin ⁡ x = 5 \sin x = 5 are possible solutions. However − 1 ≤ sin ⁡ x ≤ 1 -1 \leq \sin x \leq 1 , so sin ⁡ x = 5 \sin x = 5 is not possible. Therefore there is only solution and S = 1 2 S = \frac{1}{2} , hence a + b = 1 + 2 = 3 a + b = 1 + 2 = 3 .

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