A geometry problem by Sabhrant Sachan

Geometry Level 2

8 1 sin ⁡ 2 x + 8 1 cos ⁡ 2 x = 30 \large 81^{\sin^2 x} + 81^{\cos ^2 x} = 30

Which of the following values is a possible value of x x satisfying the equation above?

π 4 \dfrac{\pi}4 π 3 \dfrac{\pi}3 π 2 \dfrac{\pi}2 π 15 \dfrac{\pi}{15} 7 π 18 \dfrac{7\pi}{18} 5 π 18 \dfrac{5\pi}{18}

Sabhrant Sachan by Sabhrant Sachan , 21, India

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1 solution

Sabhrant Sachan
May 1, 2016

Multiply Both sides by 8 1 c o s 2 x ⟹ 8 1 s i n 2 x + c o s 2 x + 8 1 2 c o s 2 x = 30 × 8 1 c o s 2 x = 81 + 8 1 2 c o s 2 x = 30 × 8 1 c o s 2 x Put 8 1 c o s 2 x = y ⟹ 81 + y 2 = 30 y = y 2 − 30 y + 81 = 0 = y 2 − 27 y − 3 y + 81 = 0 = ( y − 27 ) ( y − 3 ) = 0 ⟹ 8 1 c o s 2 x = 27 , 3 = 3 4 c o s 2 x = 3 3 , 3 1 c o s 2 x = 3 4 , 1 4 x = π 6 , π 3 \text {Multiply Both sides by } 81^{cos^2x} \\ \implies 81^{sin^2x+cos^2x}+81^{2cos^2x}=30\times81^{cos^2x}\\ = 81+81^{2cos^2x}=30\times81^{cos^2x}\\ \text {Put } 81^{cos^2x}=y \\ \implies 81+y^2=30y \\ =y^2-30y+81=0 \\= y^2-27y-3y+81=0\\ = (y-27)(y-3)=0 \\\implies 81^{cos^2x}=27,3 \\ = 3^{4cos^2x}=3^3,3^1\\ cos^2x=\dfrac34,\dfrac14\\ x=\dfrac{\pi}6,\boxed{\dfrac{\pi}3}

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Pretty much got the same solution, except I replaced sin(x) with 1 - cos^2(x) and performed the same substitution after...

Jiahui Tan - 1 year, 9 months ago

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