Question

a)sin4x+cos^2×2x=2
б)3sin^2x+sin2x=2

Answer 1

3sin²x+sin2x=2⇒3sin²x+2sinxcosx-2sin²x-2cos²x=0
sin²x+2sinxcosx-2cos²x=0⇒делим на cos²x:
tg²x+2tgx-2=0;⇒tgx=y;
y²+2y-2=0;⇒y=-1⁺₋√1+2=-1⁺₋√3;
y₁=-1+√3;⇒tgx=-1+√3=0.73;⇒x=arctg0.73+kπ;k∈Z;
y₂=-1-√3;⇒tgx=-1-√3=-2.73;⇒x=arctg(-2.73)+kπ;k∈Z.