Question

Решите что нибудь?!!!!!

Answer 1

b1. cosx = -sqrt (1 - 25/169) = -sqrt(144/169) = -12/13

b2. 
(sinx)^1 + (cosx)^2 + (tgx)^2 = 1 + (tgx)^2 = 1 / (cosx)^2

b3.
2sinx - 1 = 0
2sinx = 1
sinx = 1/2
x = (-1)^k arcsin (1/2) + Pi*k, k прин Z
x = (-1)^k Pi/6 + Pi*k, k прин Z

b4. sinx + cos(2Pi+x) - cos (Pi/2-x) = -1
sinx + cosx - sinx = -1
cosx = -1
x = +- arccos(-1) + 2Pi*k, k прин Z
x = +- (Pi - arccos1) + 2Pi*k, k прин Z
x = +- (Pi - 0) + 2Pi*k, k прин Z
x = +- Pi + 2Pi*k, k прин Z

b5. 10^(1/4)*40^(1/4)*5^(1/2) = 10^(1/4)*40^(1/4)*25^(1/4) = 10000^(1/4) = 10

b6. 3^(2x-13) = 1/243
3^(2x-13) = 3^(-5)
2x - 13 = -5
2x = 8
x = 4

b7. 
4^(2-3x) >= 64
4^(2-3x) >= 4^(3)
2-3x >= 3
-3x >= 1
x <= -1/3

b8. ОДЗ: 33-3x>=0
-3x >= -33
x<= 11

sqrt(33-3x) = 6
33 - 3x = 36
-3x = 3
x = -1

b9. ОДЗ:
-7/(32-8x) > 0 | * (-1/7)
32 - 8x < 0
-8x < -32
x > 4

sqrt (-7/(32-8x)) = 1/4

-7/(32-8x) = 1/16
32 - 8x = -7*16
-8x = -112 - 32
-8x = -144
x = 18

b10. 
ОДЗ:
{-2x + 15>=0
{x >= 0

{-2x >= -15
{x>= 0

{x <= 15/2
{x>= 0

0<= x<= 15/2

sqrt(-2x+15) = x
-2x + 15 = x^2
x^2 + 2x - 15 = 0
D = 4-4*1*(-15) = 64
x1 = (-2 + 8)/2 = 6/2 = 3
x2 = -10/2 = -5 - не принадлежит ОДЗ

с1.
2(sinx)^2 + 7cosx + 2 = 0
2(1 -(cosx)^2) + 7cosx + 2 = 0
2 -2(cosx)^2 + 7cosx + 2 = 0
-2(cosx)^2 + 7cosx + 4 = 0
замена cosx = t, -1 <= t <= 1
-2t^2 +7t + 4 = 0
D = 49-4*(-2)*4 = 81
t1 = (-7+9)/-4 = -2/4 = -1/2
t1 = (-7-9)/-4 = 16/4 = 4 не принадлежит [-1;1]
cosx = -1/2
x = +- arccos(-1/2) + 2Pi*k, k прин Z
x = +- (Pi - arccos1/2) + 2Pi*k, k прин Z
x = +- (Pi - Pi/3) + 2Pi*k, k прин Z
x = +- (2Pi/3) + 2Pi*k, k прин Z