1) sin(pi/3-2x)*cos(pi/3-2x)< -√3/2 2) sin2x+√3cos2x≥-1

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вопрос

1) To solve the inequality sin(pi/3-2x)*cos(pi/3-2x) < -√3/2, we will first simplify the expression by using the sum and difference identities for sine and cosine:

sin(pi/3-2x)cos(pi/3-2x) = sin(pi/3)cos(2x) - sin(2x)*cos(pi/3
= (√3/2)(cos(2x)) - sin(2x)(1/2
= (√3/2)cos(2x) - (1/2)sin(2x)

Now we need to determine when this expression is less than -√3/2:

(√3/2)cos(2x) - (1/2)sin(2x) < -√3/
√3cos(2x) - sin(2x) < -√
sin(2x) - √3cos(2x) > √3

Therefore, the inequality sin(pi/3-2x)*cos(pi/3-2x) < -√3/2 is equivalent to sin(2x) - √3cos(2x) > √3.

2) To solve the inequality sin(2x) + √3cos(2x) ≥ -1, we will first rewrite it in the form of a single trigonometric function using the sum identity for sine:

sin(2x) + √3cos(2x) = 2sin(x)cos(x) + √3cos(2x
= 2sin(x)cos(x) + √3(2cos^2(x) - 1
= 2sin(x)cos(x) + 2√3cos^2(x) - √3

Now we need to determine when this expression is greater than or equal to -1:

2sin(x)cos(x) + 2√3cos^2(x) - √3 ≥ -
2sin(x)cos(x) + 2√3cos^2(x) ≥ -
sin(2x) + 2√3(cos^2(x) - sin^2(x)) ≥ -2

Therefore, the inequality sin(2x) + √3cos(2x) ≥ -1 is equivalent to sin(2x) + 2√3(cos^2(x) - sin^2(x)) ≥ -2.