Question

precalculus question love help!!!

Answer 1

4sin(x) + 4cos^2(x) = 4
sin(x) + cos^2(x) = 1
sin(x) + 1-sin^2(x) = 1
sin^2(x) -sin(x) = 0
sin(x) [ sin(x) -1 ] = 0
x = 0, pi/2

Answer 2

lol... use sin^2 + cos^2 = 1

then cos^2 = 1 - sin^2

substitute it into your equation

4sin(x) + 4(1 - sin^2(x)) = 4

divide by 4

sin(x) + 1 - sin^2(x) = 1

or sin(x) - sin^2(x) = 0

factorise

sin(x)(1 - sin(x)) = 0

solutions are when sin(x) = 0 or 1

gee I think I've done this question before

Answer 3

so is the solution 0, pi/2

or 0, 1

Answer 4

and thanks for all your help today campbell

Answer 5

The answers occur when sin(x) is 0, 1
The answers ARE 0, pi/2

Answer 6

the domain for t is [0, 2pi] so you need to include pi and 2pi... as sin(t) = 0 at those points...

but I'm not sure on the answer restrictions

Answer 7

it actually says 0,pi/2 arent correct so are those the answers for just throughout that first rotation? because the domain is 2pi are there two more solutions
?

Answer 8

the way I read the question, there are 4 answers, 0, pi/2, pi, 2pi

Answer 9

[0, 2pi) means that 2pi cannot be included in the answer choice set

Answer 10

oh shoot yeah i forgot about pi

Answer 11

So the answers are 0, pi/2, pi

Answer 12

ok thanks wombat... so exclude 2pi... its 0, pi/2 and pi

Answer 13

YALL ROCK!