Question

Решите уравнение:
$16 {}^{ \sin {}^{2}x } + 16 {}^{ \cos {}^{2} x } = 10$
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Answer 1

16^(sin^2x)+16^(1-sin^2x)=10

16^(sin^2x)=t

t+16/t=10

t^2-10t+16=0

D=100-64=6^2

t=(10+6)/2=8; 16^(sin^2x)=8; 2^(4sin^2x)=2^3; 4sin^2x=3; sin^2x=3/4;

sinx=+-√3/2; x=+-pi/3+pik

t2=(10-6)/2=2; 2^(4sin^2x)=2^1; 4sin^2x=1; sin^2x=1/4; sinx=+-1/2; x=+-pi/6+pin

k,n∈Z