Soviet Trig Eqn. 1

Geometry Level 5

Solve, for 0 ∘ < x < 9 0 ∘ 0^{\circ} < x < 90^{\circ} , the following equation for x x :

13 − 12 cos ⁡ x + 7 − 4 3 sin ⁡ x = 2 3 \sqrt{13-12\cos x}+\sqrt{7-4\sqrt{3}\sin x}=2\sqrt{3}

If x x can be represented in the form arccos ⁡ ( a + b c ) \arccos\left( \dfrac{a+\sqrt{b}}{c}\right) , find a + b + c a+b+c .


The answer is 32.

Hobart Pao by Hobart Pao , 23, USA

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1 solution

Tanishq Varshney
Aug 27, 2015

let cos ⁡ x = c \cos x=c and sin ⁡ x = s \sin x=s , so s 2 + c 2 = 1 s^2+c^2=1

13 − 12 c = 2 3 − 7 − 4 3 s \large{\sqrt{13-12c}=2\sqrt{3}-\sqrt{7-4\sqrt{3}s}}

squaring and simplifying

2 3 s − 6 c − 3 = − 2 3 7 − 4 3 s \large{2 \sqrt{3}s -6c-3=-2 \sqrt{3} \sqrt{7-4\sqrt{3}s}}

squaring both sides we get

12 s 2 + 36 c 2 + 9 − 24 3 s c − 12 3 s + 36 c = 84 − 48 3 s \large{12s^2+36c^2+9-24\sqrt{3}sc-12\sqrt{3}s+36c=84-48\sqrt{3}s}

12 ( 1 − c 2 ) + 36 ( 1 − s 2 ) + 9 − 24 3 s c − 12 3 s + 36 c = 84 − 48 3 s \large{12(1-c^2)+36(1-s^2)+9-24\sqrt{3}sc-12\sqrt{3}s+36c=84-48\sqrt{3}s}

12 c 2 + 36 s 2 + 27 + 24 3 s c − 36 3 s − 36 c = 0 \large{12c^2+36s^2+27+24\sqrt{3}sc-36\sqrt{3}s-36c=0}

( 2 3 c + 6 s − 3 3 ) 2 = 0 \large{(2\sqrt{3}c+6s-3\sqrt{3})^2=0}

2 3 c − 3 3 = − 6 s \large{2\sqrt{3}c-3\sqrt{3}=-6s}

squaring

12 c 2 + 27 − 36 c = 36 ( 1 − c 2 ) \large{12c^2+27-36c=36(1-c^2)}

16 c 2 − 12 c − 3 = 0 \large{16c^2-12c-3=0}

c = 3 + 21 8 \large{\boxed{c=\frac{3+\sqrt{21}}{8}}}

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