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2)
![2 (sin(x))^{2} + 3 cos(x) = 0 2 (sin(x))^{2} + 3 cos(x) = 0](https://tex.z-dn.net/?f=2+%28sin%28x%29%29%5E%7B2%7D+%2B+3+cos%28x%29+%3D+0)
![2(1 - { (cos(x)) }^{2} ) + 3 cos(x) = 0 2(1 - { (cos(x)) }^{2} ) + 3 cos(x) = 0](https://tex.z-dn.net/?f=2%281+-+%7B+%28cos%28x%29%29+%7D%5E%7B2%7D+%29+%2B+3+cos%28x%29+%3D+0)
![- 2( cos(x) )^{2} + 3 cos(x) + 2 = 0 - 2( cos(x) )^{2} + 3 cos(x) + 2 = 0](https://tex.z-dn.net/?f=+-+2%28+cos%28x%29+%29%5E%7B2%7D+%2B+3+cos%28x%29+%2B+2+%3D+0)
![2( { cos(x)) }^{2} - 3 cos(x) - 2 = 0 2( { cos(x)) }^{2} - 3 cos(x) - 2 = 0](https://tex.z-dn.net/?f=2%28+%7B+cos%28x%29%29+%7D%5E%7B2%7D+-+3+cos%28x%29+-+2+%3D+0)
Пусть cosx = t
![2 {t}^{2} - 3t - 2 = 0 2 {t}^{2} - 3t - 2 = 0](https://tex.z-dn.net/?f=2+%7Bt%7D%5E%7B2%7D+-+3t+-+2+%3D+0)
D = 9 - 4*2*(-2) = 25
t1 = (3+5)/4 = 2
t2 = (3-5)/4 = -1/2
cosx = 2
x= +- arccos2 + 2*pi*n, n€Z
cosx = -1/2
x = +- (pi - arccos1/2) + 2*pi*n, n€Z
3) ctgx + 3tgx = 2 sqrt (3)
1/tgx + tgx = 2sqrt (3)
(tg^2x + 1)/tgx = 2sqrt (3)
tg^x - 2sqrt (3)*tgx +1 =0
Пусть tgx=t
t^2-2sqrt (3)t+1 =0
D = 12 - 4*1 = 8
t1,2 = (2sqrt (3)+-2sqrt (2))/2
x = arctg(sqrt (3)+sqrt (2)) + pi*n, n €Z
x= arctg (sqrt (3)-sqrt (2)) + pi*n, n €Z
4) tgx = ctgx
tgx/ctgx = 1
tg^2x = 1
tgx=1
tgx=-1
x=pi/4 + pi*n, n €Z
x= -pi/4 + pi*n, n €Z
5 и 6 не уверен, как решать.
Пусть cosx = t
D = 9 - 4*2*(-2) = 25
t1 = (3+5)/4 = 2
t2 = (3-5)/4 = -1/2
cosx = 2
x= +- arccos2 + 2*pi*n, n€Z
cosx = -1/2
x = +- (pi - arccos1/2) + 2*pi*n, n€Z
3) ctgx + 3tgx = 2 sqrt (3)
1/tgx + tgx = 2sqrt (3)
(tg^2x + 1)/tgx = 2sqrt (3)
tg^x - 2sqrt (3)*tgx +1 =0
Пусть tgx=t
t^2-2sqrt (3)t+1 =0
D = 12 - 4*1 = 8
t1,2 = (2sqrt (3)+-2sqrt (2))/2
x = arctg(sqrt (3)+sqrt (2)) + pi*n, n €Z
x= arctg (sqrt (3)-sqrt (2)) + pi*n, n €Z
4) tgx = ctgx
tgx/ctgx = 1
tg^2x = 1
tgx=1
tgx=-1
x=pi/4 + pi*n, n €Z
x= -pi/4 + pi*n, n €Z
5 и 6 не уверен, как решать.
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