Forum Gusht 2018.docx

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1. Zgjidhni 𝑠𝑖𝑛2π‘₯ = 𝑠𝑖𝑛π‘₯ + π‘π‘œπ‘ π‘₯ cos 2 π‘₯ βˆ’ sin2 π‘₯ βˆ’ (𝑠𝑖𝑛π‘₯ + π‘π‘œπ‘ π‘₯) = 0 => (𝑠𝑖𝑛π‘₯ + π‘π‘œπ‘ π‘₯)(π‘π‘œπ‘ π‘₯ βˆ’ 𝑠𝑖𝑛π‘₯π‘₯) βˆ’ (𝑠𝑖𝑛π‘₯ + π‘π‘œπ‘ π‘₯) = 0 (𝑠𝑖𝑛π‘₯ + π‘π‘œπ‘ π‘₯)(π‘π‘œπ‘ π‘₯ βˆ’ 𝑠𝑖𝑛π‘₯ βˆ’ 1) = 0 3πœ‹ 𝑠𝑖𝑛π‘₯ + π‘π‘œπ‘ π‘₯ = 0 => π‘‘π‘Žπ‘›π‘₯ = βˆ’1 => π‘₯ = π‘˜πœ‹ + 4 1 1 1 π‘π‘œπ‘ π‘₯ βˆ’ 𝑠𝑖𝑛π‘₯ βˆ’ 1 = 0 => π‘π‘œπ‘ π‘₯ βˆ’ 𝑠𝑖𝑛π‘₯ = βˆ’1 => π‘π‘œπ‘ π‘₯ βˆ’ 𝑠𝑖𝑛π‘₯ = √2 √2 √2 πœ‹ πœ‹ 1 cos π‘π‘œπ‘ π‘₯ βˆ’ sin 𝑠𝑖𝑛π‘₯ = 4 4 √2 πœ‹ 1 cos (π‘₯ + ) = 4 √2 πœ‹ πœ‹ π‘₯ + = 2π‘˜πœ‹ Β± 4 4 2. gjeni siperfaqen e trekendeshit BED.

3. ABCDEF 6- kendesh I rregullt . AK=KF; BC=LC; AB=x dhe KL=9-x. Gjeni 5x=?

4. Zgjidhni |π‘₯ βˆ’ 3| < π‘₯ + 1; 5. zgjidhni ne R.

]1; +∞[ ] βˆ’ ∞; βˆ’3] βˆͺ] βˆ’ 1; 2[βˆͺ [3,5] 3π‘₯ βˆ’ 9 ≀0 𝑓(π‘₯)

6. zgjidhni ne R. βˆ’

𝑓(π‘₯) <0 π‘₯2 βˆ’ 9

7. Trekendeshi EBC barabrinjes. Gjeni kendin a.

8. ABCD drejtekendesh. EF=FB=2cm, FC=4cm. Gjeni siperfaqen e ABCD=?

9. Gjeni AB=x

𝐴𝐡𝐷:

π‘Ž π‘₯ = ; 𝑠𝑖𝑛90 𝑠𝑖𝑛72

π‘Ž 12 π‘₯𝑠𝑖𝑛36 12𝑠𝑖𝑛18𝑠𝑖𝑛72 = ; 𝑠𝑖𝑛18 = => π‘₯ = 𝑠𝑖𝑛18 𝑠𝑖𝑛36 12𝑠𝑖𝑛72 𝑠𝑖𝑛36 1 12 βˆ— 2 [π‘π‘œπ‘ 54 βˆ’ π‘π‘œπ‘ 90] = 6π‘π‘š. 𝑠𝑖𝑛36

𝐷𝐸𝐢:

1. Gjeni kendin a.

2. Gjeni kendin a.

3. Gjeni gjatesine BE.

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