19-mavzu. Trigonometrik tenglama va tengsizliklarni o‘qitish metodikasi Reja

4-misol. sin2x+3cos2x=a tenglama yechilsin.
yechish. 2sinxcosx+3cos²x-3sin²x=a,
2tgx+3–3tg²x=a(1+tg²x),
tg²x(a+3)–2tgx+(a–3)=0,
.
Agar |a| bo’lsa, Agar a=–3 bo’lsa, sin2x+3cos2x=–3 bo’ladi.
2sinxcosx+3cos²x+3cos²x–3=–3, 2sinxcosx+6cos²x=0, cosx(sinx+3cosx)=0,
1) cosx=0, sinx+3cosx0, x₁= +k, kz;
2) cosx0, sinx+3cosx=0, x₂=arctg(–3)+k, kz.
J: Agar |a| (a–3) bo’lsa, , agar (a–3) bo’lsa, x= +k, kz; agar |a|> bo’lsa, tenglama yechimga ega emas.
5-misol. sin⁶x+cos⁶x=a(sin⁴x+cos⁴x) tenglama yechilsin.
yechish. sin⁴x+cos⁴x=(sin²x+cos²x)²–2sin²xcos²x=1–2sin²xcos²x,
sin⁶x+cos⁶x=(sin²x+cos²x)³–3sin⁴xcos²x–3sin²x·cos⁴x)=1–3sin²xcos²x,
1–3sin²xcos²x=a(1–2sin²xcos²x),
1–3sin²xcos²x=a–2asin²xcos²x,
sin²xcos²x= , sin²2x=4 .
Agar bu yerda 04 1 bo’lsa, tenglama ma’noga ega bo’ladi. Bu tengsizlikni yechsak, a1 bo’ladi. Agar a= bo’lsa, 1–3sin²xcos²x= –3sin²xcos²x hosil bo’ladi, bundan 1= tenglik hosil bo’ladi, buning bo’lishi mumkin emas.
Agar a1 bo’lsa,
Agar a< va a>1 bo’lsa, yechim yo’q.
6-misol. tenglama yechilsin.
yechish. bunga ko’ra




bu yerda quyidagi shartlar bajarilishi kerak:

Bu tengsizliklarni yechsak, hosil bo’ladi. Bu shartga ko’ra

Buni yechsak,
J: Agar bo’lsa, x₁=2k, kz
Agar 1m5 bo’lsa, x₁=2k, kz
Agar bo’lsa, x=2k, kz .
7-misol. asin2x+bsin + tenglamani yeching.
yechish. Bizga ma’lumki, asinx+bcosx=Asin(x+).
Bu yerda A= burchakning qiymati esa quyidgai shartlardan kelib chiqadi: sin= , cos= , =arctg . Shuning uchun sin(2x–)–sin6x=0, sin cos =0;

1. 
   sin =0, 4x+=(–1)^karcsin0⁰+2k,
   

4x=–+2k, x= – arctg .
b) cos =0, =arccos0⁰+2k, 8x–=+2k.
J: x= + arctg , kєz.

MUSTAQIL YEChISh UChUN MISOLLAR.

1-misol. sesx+cosesx+sesxcosesx=a, (a0) tenglama yechilsin.
J: Agar 1 bo’lsa, x=– +(–1)^karcsin +k, kєz.
2-misol. tg(a+x)tg(a–x)=1–2cos2x tenglama yechilsin.
J: x=k arccos