Question

решите,пожалуйста

 

3cos2x+10sin*2x-6=0

Answer 1

3cos2x+10sin*2x-6=0

3(cos^2x-sin^2x)+20sinxcosx-6(sin^2x+cos^2x)=0

3cos^2x-3sin^2x+20sinxcosx-6sin^2x-6cos^2x=0

9sin^2x-20sinxcosx+3cos^2x=0   /:cos^2x≠0

9tg^2x-20tgx+3=0

tgx=t

D=400-108=292

t=(20+2√73)/18=(10+√73)/9

t=(20-2√73)/18=(10-√73)/9

 

tgx=(10+√73)/9

x=arctg(10+√73)/9+pik, k∈Z

 

tgx=(10-√73)/9

x=arctg(10-√73)/9+pik, k∈Z