4*(1-cos2x)/2 *(1+cos2x)=1-cos2x
2*(1-cos2x) *(1+cos2x)-(1-cos2x)=0
(1-cos2x)(2+2cos2x-1)=0
(1-cos2x)(2cos2x+1)=0
1-cos2x=0⇒cos2x=1⇒2x=2πn⇒x=πn
cos2x=-1/2⇒2x=+-2π/3+2πn⇒x=+-π/3+πn
<span>Log3 (2x-5)+log3 (2x-3)=log3(3)
(2x-5)(2x-3)=3
4x^2+15-16x=3
4x^2-16x+12=0
x=[8+-sqrt(64-48)]/4=[8+-4]/4
x=1
x=3
2x-5>= x>2,5
оответ х=3
</span>
Решение
1) cos720+tg 30 ctg 210+sin 120 = сos2*360° + tg30°ctg(180° + 30°) +
+ sin(90° + 30)° = 1 + 1/√3*ctg30° + sin30° = 1 + (1/√3)*(√3) + 1/2 = 1 + 1 + 1/2 = 2(1/2)
<span>tg0 </span>°<span>- 2ctg90</span>° <span>- sin0</span>° <span>- 3cos90</span>° = 0 - 2*0 - 0 - 3*0 = 0