Question

Please help with both precalc questions! 2 dif problems

Answer 1

Explanation:

For the first problem you have:

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

Factor out a sin(x) to get:

`sin(x)[2sin(x)-1]=0`

Individually, set both terms equal to zero:

`sin(x)=0`

`2sin(x)-1-0`

The first equation is asking "sine of what angle gives zero?" The sine of 0 will give zero, and so will the sine of pi, and 2pi, etc. Therefore, you can generalize this by saying the solution is pi multiplied by an integer 'n' where 'n' can be from zero to infinity:

`x=n\pi`

where n = 0, 1, 2, 3, 4, 5, etc..

(make sure you specify that 'n' can equal zero.)

For the second equation, add 1 to both sides to get:

`2sin(x)=1`

Divide both sides by 2 to get:

`sin(x)=(1)/(2)`

Sine of what gives 1/2? Sine of pi/6 gives 1/2, and so does 5pi/6. Therefore, the two solution here would be pi/6 plus a phase shift of 2pi, and 5pi/6 plus a phase shift of 2pi:

`x=(\pi )/(6) +2n\pi`

`x=(5\pi )/(6) +2n\pi`

The second problem states that:

`cos(u)=(4)/(5)`

Cosine is defined as the adjacent side of a right triangle divided by the hypotenuse of the right triangle:

`cos(u)=adjacent/hypotenuse`

After drawing a triangle, the adjacent side is 4, and the hypotenuse is 5. But first, lets rewrite sin(2u) using the double angle identity to get:

`sin(2u)=2sin(u)cos(u)`

But we know the value of cosine of 'u', it is 4/5. Therefore:

`sin(2u)=2sin(u)*(4)/(5)`

`sin(2u)=(8)/(5) sin(u)`

To find the sine of 'u', use the triangle you should have constructed when I defined cosine as the adjacent over the hypoptenuse. Sine is defined as the opposite over the hypotenuse. We have:

`sin(u)=opposite/hypotenuse`

We have the adjacent, we have the hyppotenuse, and we want the opposite. If we find the opposite side, we can find the sin(2u). Use pythagogrean's theorem to find the opposite:

`adjacent^2+opposite^2=hypotenuse^2`

`4^2+opposite^2=5^2`

`opposite^2=25-16`

`opposite=3`

Finally, we have that:

`sin(u)=(3)/(5)`

Or:

`sin(2u)=(8)/(5)*(3)/(5) =24/25`

Hopefully you could follow along. There is a lot going on here, but since youre in pre-calc, you should be able to follow along as long as you follow my steps.