Question

2sin2x-2cos2x-√3=0
Решите уравнение

Answer 1

$2sin2x-2cos2x- sqrt{3} =0 \ 2 (2sinxcosx-cos^2x+sin^2x)-sqrt{3} =0\ 4sinxcosx-2cos^2x+2sin^2x-sqrt{3}(cos^2x+sin^2x) =0\ 4sinxcosx-(2+sqrt{3})cos^2x+(2-sqrt{3})sin^2x=0\ 4tgx-(2+sqrt{3})+(2-sqrt{3})tg^2x=0\ (2-sqrt{3})t^2+4t-(2+sqrt{3})=0 \ D=16+4*(2-sqrt{3})*(2+sqrt{3}) = 16+4(4-3)=16+4=20\t_{1}= frac{-4+2 sqrt{5} }{2(2-sqrt{3})} = frac{-2+ sqrt{5} }{2-sqrt{3}} \ t_{2}= frac{-4-2 sqrt{5} }{2(2-sqrt{3})} = frac{-2- sqrt{5} }{2-sqrt{3}} \ tgx = frac{-2+ sqrt{5} }{2-sqrt{3}} \x = arctg(frac{-2+ sqrt{5} }{2-sqrt{3}}) + pi k, k in Z\tgx = frac{-2- sqrt{5} }{2-sqrt{3}} \x = arctg(frac{-2- sqrt{5} }{2-sqrt{3}}) + pi n, k n Z$

Answer 2

4sinxcosx-2cos²x+2sin²x-√3cos²x-√3sin²x=0
(2-√3)*sin²x+4sinxcosx-(2+√3)*cos²x=0/cos²x
(2-√3)tg²x+4tgx-(2+√3)=0
tgx=a
(2-√3)*a²+4a-(2+√3)=0
D=16+4(2-√3)(2+√3)=16+4*1=20
a1=(-4-2√5)/(4-2√3)=(-2-√5)/(2-√3)⇒tgx=(-2-√5)/(2-√3)⇒
x=arctg[(2+√5)/(√3-2)]+πk,k∈z
a2=(-2+√5)/(2-√3)⇒tgx=(√5-2)/(2-√3)⇒x=arctg[((√5-2)/(2-√3)]+πk,k∈z