Question

question will be attached plz help if you can :)

Answer 1

Which question?

Answer 2

@SantanaG Can you clearly specify which question(s) you need help?

Answer 3

any @finaiized and @Callisto

Answer 4

i dont understand how to do any of the problems so i will take any help plz

Answer 5

The first step is to make sure each question uses the same trigonometric ratio.

Answer 6

I will do 10.
\[-\sin^2x=2\cos x-2\]
We want the same trigonometric ratios. So use the Pythagorean identity to convert the \(-\sin^2x\) to \(\cos^2x-1\)

\[\cos^2x-1=2\cos x-2\]
\[\cos^2x-2\cos x+1=0\]

Hopefully you can solve the remaining bit.

Answer 7

the glass steagall act regulated act regulated he stock market t or f

Answer 8

@ilovejake224 ask that in the history section

Answer 9

Alright, then I do 11
\[2sin(2x) -2sinx + 2\sqrt3cosx -\sqrt3 =0\]\[4sinxcosx -2sinx + 2\sqrt3cosx -\sqrt3 =0\]\[2sinx(2cosx -1) + \sqrt3(2cosx -1) =0\]\[(2sinx+\sqrt3)(2cosx -1) =0\]\[(2sinx+\sqrt3)=0 \ or \ (2cosx -1) =0\]
I think you can solve it!

Answer 10

For 12,
csc(4x) -2=0
\[\frac{1}{sin4x} =2\]\[sin4x =\frac{1}{2}\]
Now, you can solve it o.o...

Answer 11

For 13,
\[cos^2(2x)-sin^2(2x) =0\]\[cos[2(2x)] =0\]\[cos(4x) =0\] Your turn to work it out!

Answer 12

or 11. x=0 or x=1/396 pi)

Answer 13

and 10.x=2pi?

Answer 14

For the last one
sin2x = sinx
2sinxcosx =sinx
2cosx = 1
What's next?

Answer 15

hold on im still on 12 lol :3

Answer 16

11 => no

Answer 17

12. x=pi/24 ?

Answer 18

10 => yes, but you've missed another solution..

Answer 19

11. x=0 and x=1/3 (6pi)

Answer 20

@Callisto - Sorry for wasting your effort when you were doing 10! :(

Answer 21

12. => yes, but you've missed another solution (in quad. II)

Answer 22

@finaiized Never mind :)

Answer 23

11=> no

Answer 24

idk what to do then for 11. sorry

Answer 25

Consider,
\[2sinx+\sqrt3 =0\]\[sinx=-\frac{\sqrt3}{2}\] Solve x
Consider
\[2cosx-1=0\]\[cosx=\frac{1}{2}\]Solve x
Plug in the answers you've got into the equation again, reject if it doesn't make sense :)

Answer 26

For the above, make sure you recognize that they are special angles. This is how you can solve exactly. Also watch the negatives to check which quadrants they are in using CAST.

Answer 27

ok thank you @Callisto and @finaiized

Answer 28

welcome :)