solve the following equations algebraically for 0< x < 2pi

Question

2 tanx -3 =0

 

3 cos^2 x- 2sinx= -2

 

csc^2x + 2cotx -5 = 0

Answer 1

2tan(x)-3=0, tan(x)=3/2=1.5, x=arctan(1.5)=0.9828 radians approx. Also, x=0.9828+π=4.1244 radians.

3cos²(x)-2sin(x)=-2, 3(1-2sin²(x))-2sin(x)=-2, 3-6sin²(x)-2sin(x)=-2, 6sin²(x)+2sin(x)=5, sin²(x)+⅓sin(x)=⅚, sin²​(x)+⅓sin(x)+1/36=⅚+1/36, (sin(x)+⅙)²=31/36, sin(x)+⅙=±√31/6, sin(x)=-⅙±√31/6, sin(x)=0.7613 approx. (sin(x) cannot be less than -1), x=0.8653 or π-0.8653=2.2763 radians.

Note that since sin²(x)+cos²(x)≡1, 1+cot²(x)≡csc²(x).

csc²(x)+2cot(x)-5=0, 1+cot²(x)+2cot(x)-5=0, (cot(x)+1)²=5, cot(x)+1=±√5, cot(x)=-1±√5, tan(x)=¼(1+√5) or ¼(1-√5), x=0.6802 or -0.2997. Also, x=0.6802+π=3.8218. For negative x we can replace with 2π+x, so -0.2997=2π-0.2997=5.9835 radians approx; also, x=5.9835-π=2.8419 radians.