Решите уравнение:
а)3(cos^2)x +1/2sinx=2
б)2(sin^2)x - 5sinx*cosx-(cos^2)x=-2
a) 3cos^2x+1/2-1/2cos^2x=2
2.5cos^2x=1,5
cos^2x=0,6
1+cos2x=1,2
cos2x=0,2
2x=+/-arccos0,2+2pn
x=+/-arccos0,2/2+pn
b)2tg^2x-5tgx+1=0
2a^2-5a+1=0
d=25-8=17
a1,2=5+/-sgrt17/4
[tgx=1,25+sgrt17/4 [x=arctg1,25+sgrt17/4 +pn
[tgx=1,25-sgrt17/4 [x=arctg1,25-sgrt17/4 + pn
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Answers & Comments
a) 3cos^2x+1/2-1/2cos^2x=2
2.5cos^2x=1,5
cos^2x=0,6
1+cos2x=1,2
cos2x=0,2
2x=+/-arccos0,2+2pn
x=+/-arccos0,2/2+pn
b)2tg^2x-5tgx+1=0
2a^2-5a+1=0
d=25-8=17
a1,2=5+/-sgrt17/4
[tgx=1,25+sgrt17/4 [x=arctg1,25+sgrt17/4 +pn
[tgx=1,25-sgrt17/4 [x=arctg1,25-sgrt17/4 + pn