1) (x-2)(x-1) = 0x = 2 or x = 1
2) Replace sinα and cosα in the given expression:sinα - 4cosα / 2sinα + cosα = sinα - 4√(1 - sin²α) / 2sinα + cosα= sinα - 4√(1 - sin²α) / 2sinα + √(1 - sin²α)
Given that tg(α/2) = 2, we can find sinα and cosα:sinα = 2/(1+2²) = 2/5cosα = √(1 - (2/5)²) = √(21)/5
Substitute sinα and cosα back into the expression:(2/5) - 4(√(21)/5) / 2(2/5) + √(21)/5= (2/5) - 4(√21)/5 / (4/5) + √21/5= (2 - 4√21) / 4 + √21= (2 - 4√21) / 4 + √21
3) Simplify the expression:3(log₂3 + log₃16 + 4)(log₂3 - 2log₁₂3)log₃2 - log₂3= 3(log₂3 + 4 + 2log₃2)(log₂3 - log₂3²) - log₂3= 3(log₂3 + 4 + 2)(log₂3 - 3) - log₂3= 3(4 + log₂3 + 8)(log₂3 - 3) - log₂3= 3(12 + log₂3)(log₂3 - 3) - log₂3
4) y = 1/√(x² - 4)This is the equation of a hyperbola:
5) 1 + cosα + cos2α + cos3α / sin2α + 2sinαcos2α = ctg(α)To simplify the expression, use trigonometric identities and properties of cotangent.
1) (x-2)(x-1) = 0
x = 2 or x = 1
2) Replace sinα and cosα in the given expression:
sinα - 4cosα / 2sinα + cosα = sinα - 4√(1 - sin²α) / 2sinα + cosα
= sinα - 4√(1 - sin²α) / 2sinα + √(1 - sin²α)
Given that tg(α/2) = 2, we can find sinα and cosα:
sinα = 2/(1+2²) = 2/5
cosα = √(1 - (2/5)²) = √(21)/5
Substitute sinα and cosα back into the expression:
(2/5) - 4(√(21)/5) / 2(2/5) + √(21)/5
= (2/5) - 4(√21)/5 / (4/5) + √21/5
= (2 - 4√21) / 4 + √21
= (2 - 4√21) / 4 + √21
3) Simplify the expression:
3(log₂3 + log₃16 + 4)(log₂3 - 2log₁₂3)log₃2 - log₂3
= 3(log₂3 + 4 + 2log₃2)(log₂3 - log₂3²) - log₂3
= 3(log₂3 + 4 + 2)(log₂3 - 3) - log₂3
= 3(4 + log₂3 + 8)(log₂3 - 3) - log₂3
= 3(12 + log₂3)(log₂3 - 3) - log₂3
4) y = 1/√(x² - 4)
This is the equation of a hyperbola:
5) 1 + cosα + cos2α + cos3α / sin2α + 2sinαcos2α = ctg(α)
To simplify the expression, use trigonometric identities and properties of cotangent.