Question

решите уравнение:
4sin2x+3cos2x=5

Answer 1

4sin2x + 3cos2x=5
4*2sinxcosx + 3(cos²x- sin²x)=5(cos²x + sin²x)
8sinxcosx + 3cos²x- 3sin²x - 5cos² -5sin²x = 0 
- 8sin²x  + 8sinxcosx - 2cos²x = 0
(- 8sin²x  + 8sinxcosx - 2cos²x ) / - cos²x =0
8tg²x - 8tgx + 2=0  x≠ π/2 + πn n∈ Z
tgx= y  
8у² - 8у + 2=0
d=64- 4*2*8 = 64 - 64 =0
у = -8/ 16= - 1/2
tgx=-1/2 
x = arctg(-1/2) + πn n∈Z