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