MATH Q

I only can solve that 1), 2)

1)

tan x+2cot x=3

tan x+2/tan x=3

tan² x+2=3tan x

tan² x-3tan x+2=0

tanx = [3±√(9-4*1*2)]/2

tanx = [3±1]/2

tanx = 1 or 2

x = arctan 1 or arctan 2

= pi/4 or arctan 2

2)

sin² x+2sin x cos x=3cos² x

sin² x+2sin x cos x+cos² x=4cos² x

(sin x + cos x)² = (2cos x)²

sin x + cos x = 2cos x

sin x = cos x

sin² x = cos² x

sin² x + cos² x = 2cos² x

2cos² x = 1

cos x = 1/√2

x = arccos 1/√2

x = pi/4

Do you mind if I solve question 3 and 4 only?

(3) 7sin x cos x+cos² x=2

7sin x cos x + cos² x = 2 (sin² x + cos² x)

7sin x cos x + cos² x = 2 sin² x + 2 cos² x

7sin x cos x - 2 sin² x - cos² x = 0

Dividing by cos² x,

7 tan x - 2 tan² x - 1 = 0

2 tan² x- 7 tan x +1 = 0

tan x = { 7+ √(-7)² - 4(2)(1) }/4 or { 7- √(-7)² - 4(2)(1) }/4

tan x ={ 7+√41}/4 or {7-√41}/4

Therefore x = 1.28 , 4.42 , 0.148, 3.29

4)sin x-√3 cos x=√2

(sin x-√3 cos x)² = 2

sin² x - 2√3sinxcos x + 3 cos² x = 2( sin ² x + cos² x)

sin² x - 2√3sinxcos x + 3 cos² x = 2 sin ² x + 2 cos² x

sin² x + 2√3 sinxcos x- cos² x = 0

Dividing by cos² x,

tan² x + 2√3 tan x -1=0

tan x = {-2√3 +√(2√3)²-4(1)(-1)} over 2(1) or {-2√3 -√(2√3)²-4(1)(-1)} over 2(1)

tan x = {-2√3 + 4 }/ 2 or {-2√3 - 4 }/ 2

tan x = -√3 +2 or -√3 -2

x = 0.262(rejected), 3.40 or x=1.83, 4.97 (rejected)

x= 3.40 or 1.83

Remember to do the answer checking after you have squared both sides.

1)tan x+2cot x=3
tan x+2 1/tan x=3
(tan x)2+2=3tan x
(tan x-2)(tan x-1)=0
解得tan x=1或tan x=2(舍去)
2)sin² x+2sin x cos x=3cos² x