Germany | Can You Solve This??? | A Nice Olympiad Algebra Problem | 99% Failed This...

[@DanToomey-j8n]

2 and fun vod is stronger if dogma is constant. R = . Cos² x becomes pi as 2nd function. 3 graphs 1 circa.. 2 π function if x>p0. Point is f§Gunction symbol. 2π² since sinx is nominal part 3 3 becomes answer 3quotient. πr. .. I'm working

[@pspprabhat]

81^sin^2 x +81^(1-sin^2x)=30 =>m+81-81/m=30 =>m^2-30m+81=0 =>m^2-3m-27m+81=0 =>m(m-3)-27(m-3)=0 =>m=3 or 27 =>3^4sin^2×=3 &sinx=+_1/2 Also sinx=+_ _/3/2

[@oahuhawaii2141]

You didn't solve for x.

[@satrajitghosh8162]

writing z = 81 ^ (sin^ 2( x)) one gets z + 81/z = 30 z^2 - 30 z + 81 = 0 ( z - 27)( z -3) = 0 This implies z = 27, 3 = (81) ^ ( 3/4), 81 ^ (1/4) Hereby 81 ^ (sin^ 2( x)) = z = (81) ^ ( 3/4), 81 ^ (1/4) Hereby sin^ 2 ( x) = 3/4, 1/4 1 - cos (2 x) =2 sin^ 2 ( x) = 3/2, 1/2 cos (2 x) = - 1/2, 1/2 cos ^ 2 ( 2 x).= 1/4 cos (4 x) = 2 cos ^ 2 ( x) - 1 = - 1/2 Hereby 4 x = 2 n π + 2 π /3, 2 n π - 2 π /3 x = (n /2 + 1/6) π, (n /2 - 1/6) π

[@oahuhawaii2141]

You missed half the solutions. With 4*sin²(x) = 1, 3 , we have: sin(x) = ±1/2, ±√3/2 There are 8 unique solutions between -π and π .