Question

1. Create an algebraic expression for
sin(arccosx-arcsin3x)

2. The cosx=4/5, x lies in quadrant 4. Find sin x/2

3.Determine all solutions in (0,2pie) for sin4xsin^2x=3,
cot^2v(3-1)cotx=v(3), cos^2x=cosx, and
tan^2x-6tanx+4=0


4. Solve; sin(∏/2-x)=1/sec

5. Solve; cot+4x+2cot^2x+1=csc4x

6. Solve; tan(x+∏/x=(√(3)+3tanx)/(3-√(3)tanx

Answer 1

Wow! You actually want somebody to do that assignment for you, without showing any effort on your part?

I will start off with #2, then you let us know some of your steps and work

use cos 2A = 1 - 2sin^2 A
then cos x = 1 - 2sin^2 x/2
4/5 = 1 - 2sin^2 x/2

solve for sin x/2
you will end up with sin x/2 = ±1/√10
but you are told that x is in quadrant IV, so x/2 must be in quadrant II and
sin x/2 = 1/√10

Answer 2

Ok, so because sin=opp/hyp and sec=hyp/adj it will be √10/

Answer 3

no, because I solved the equation 4/5 = 1 - 2sin^2 x/2

2sin^2 x/2 = 1 - 4/5
sin^2 x/2 = 1/10
take √ of both sides
sin x/2 = ± 1/√10

why are you talking about secant?
the problem dealt only with sines and cosines.