Question

Помогите пожалуйста...
$4^{sin^2x} =(frac{1}{2} )^{sin{2}x} *4$[tex][/tex]

Answer 1

I hope this helps you

Answer 2

cosx -sinx =√2sin(π/4 - x) =0 ⇔π/4 -x= πk ⇔ x=π/4+π(-k) = π/4+πn

Answer 3

sorry? :(

Answer 4

task/29502785   Решить уравнение 4^ (sin²x) = (  (1/2) ^sin2x ) *4

Решение : 4^ (sin²x) = (  (1/2) ^sin2x ) *4 ⇔(2²)^ (sin²x) = ( (2⁻¹)^(sin2x) )*2² ⇔

(2) ^ ( 2sin²x)= ( 2)^(2 - sin2x) ⇔ 2sin²x = 2 - sin2x  ⇔2-2sin²x= sin2x⇔

2( 1- sin²x) = sin2x ⇔2cos²x = 2sinx*cosx ⇔2cosx(cosx - sinx) =0 ⇔

[ cosx = 0 ; cosx - sinx =0. [ cosx = 0 ; tgx = 1.⇔ [ x =π/2 +πn ; x =π/4+πn ,n ∈ℤ.

ответ: π/2 +πn ; π/4+πn ,n ∈ℤ .