trigonometric Question

1.
sin2x + cosx <= 0
2sinxcosx + cosx <= 0
cosx(2sinx + 1)<=0
{Case I: cosx<=0 and sinx>-1/2} or {Case II: cosx>=0 and sinx<=-1/2}
Case I: cosx <=0 => π/2<= x <= 3π/2
sinπ/6 =1/2
sin7π/6 = -1/2, when x increases further sinx becomes more negative.
So for case I the answer is π/2 <= x <= 7π/6
Case II: cosx >= 0 =>
0 <= x <= π/2<= x <= π/2 (in first quadrant) OR 3π/2 <= x <= 2π (in fourth quadrant)
Since sinx <= -1/2, x must be in the 4th quadrant.
sin11π/6 = -1/2
therefore 3π/2＜x＜11π/6
2.
Sum of roots = -k/2
Product of roots =1/2
sinx + secx = -k/2 … (1)
sinxsecx = 1/2 = tanx ...(2)
Case I: x is in first quadrant,
sinx = 1/(12+22) = 1/√5
cosx = 2/(12+22) = 2/√5
sinx + secx = 1/√5 + √5/2 = (2+5)/2√5 = 7√5/10 and k = - 7√5/5
Case II: x is in third quadrant,
sinx = - 1/(12+22) = - 1/√5
cosx = - 2/(12+22) = - 2/√5
sinx + secx = - 1/√5 - √5/2 = - (2+5)/2√5 = - 7√5/10 and k = 7√5/5
3.
(a)Sum of roots = 4/3 = sinx + cosx
Product of roots = k/3 = sinxcosx
(4/3)2 = (sinx+cosx)2
16/9 = sin2x + cos2x + 2sinxcosx
16/9 – 1 = 2k/3
7/9 = 2k/3
k = 7/6
b)
(sin x)/(1-cotx)+(cos x)/(1-tanx)
=sinx/[(sinx-cosx)/sinx] + cosx/[(cosx – sinx)/cosx]
= sin2x/(sinx – cosx) – cos2x/(sinx – cosx)
= (sinx + cosx)(sinx – cosx)/(sinx – cosx)
= sinx + cosx
=Sum of roots = 4/3

1.find the values of x in the range 0<=x<=2pi，for sin2x+cosx<=0
(the ans is 3π/2＜x＜11π/6，π/2＜x＜7π/6，I want steps)
sin2x+cosx<=0
sin2x<=－cosx
sin2x<=sin(3pi/2+x)
2sinxcosx+cosx<=0
cosx(1+2sinx)<=0
(1) cosx>=0
=>0<=x<=pi/2 or 3pi/2<=x<=2pi--------(1)
1+2sinx<=0
1<=－2sinx
－1/2>=sinx=> 7pi/6<=x<=11pi/6-------(2)
Compare (1)(2)
3pi/2<=x<=11pi/6-------------------------(3)
(2) cosx<=0 =>pi/2<=x<=3pi/2----------(4)
1+2sinx>=0
－1/2<=sinx=> 0<=x<=7pi/6 or 11pi/6<=x<=2pi--------(5)
Compare (4)(5)
pi/2<=x<=7pi/6---------------------(6)
Combine (3)(6) 3pi/2<=x<=11pi/6 or pi/2<=x<=7pi/6

2. If sin x and sec x (0＜x＜2π)are the roots of the equation 2x^2+kx+1=0, find the
value of k (leave your ans in surd)
sinx+secx=－k/2，sinxsecx=1/2
sinxsecx=1/2
tanx=1/2
(1) sinx>0，secx>0
sinx=2/sqrt(5)，cosx=1/sqrt(5)，secx=sqrt(5)
2/sqrt(5)+sqrt(5)=－k/2
2sqrt(5)/5+sqrt(5)=－k/2
k=－14sqrt(5)/5
(2) sinx<0，secx<0
sinx=－2/sqrt(5)，cosx=－1/sqrt(5)，secx=－sqrt(5)
－2/sqrt(5)－sqrt(5)=－k/2
2sqrt(5)/5+sqrt(5)=k/2
k=14sqrt(5)/5

3.the two roots of the equation 3x^2－4x+k=0 are sin x and cos x
a)find the values of k
sinx+cosx=4/3，sinxcosx=k/3
sin^2 x+2sinxcosx+cos^2 x=16/9
1+k/6=16/9
k/6=7/9
k=42/9=14/3
b)find he vales of (sin x)/(1－cotx)+(cos x)/(1－tanx)
(sin x)/(1－cotx)+(cosx)/(1－tanx)
=sin^2 x/[sinx(1－cosx/sinx)]+cos^2 x/[cosx(1－sinx/cosx)]
=sin^2x/(sinx－cosx)+cos^2x/(cosx－sinx)
=sinx+cosx=4/3