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
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
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
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
You didn't solve for x.
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
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 .