Option C:
x = 90°
Solution:
Given equation:
![\sin x=1+\cos ^{2} x](https://tex.z-dn.net/?f=%5Csin%20x%3D1%2B%5Ccos%20%5E%7B2%7D%20x)
<u>To find the degree:</u>
![\sin x=1+\cos ^{2} x](https://tex.z-dn.net/?f=%5Csin%20x%3D1%2B%5Ccos%20%5E%7B2%7D%20x)
Subtract 1 + cos²x from both sides.
![\sin x-1-\cos ^{2} x=0](https://tex.z-dn.net/?f=%5Csin%20x-1-%5Ccos%20%5E%7B2%7D%20x%3D0)
Using the trigonometric identity:![\cos ^{2}(x)=1-\sin ^{2}(x)](https://tex.z-dn.net/?f=%5Ccos%20%5E%7B2%7D%28x%29%3D1-%5Csin%20%5E%7B2%7D%28x%29)
![\sin x-1-\left(1-\sin ^{2}x\right)=0](https://tex.z-dn.net/?f=%5Csin%20x-1-%5Cleft%281-%5Csin%20%5E%7B2%7Dx%5Cright%29%3D0)
![\sin x-1-1+\sin ^{2}x=0](https://tex.z-dn.net/?f=%5Csin%20x-1-1%2B%5Csin%20%5E%7B2%7Dx%3D0)
![\sin x-2+\sin ^{2}x=0](https://tex.z-dn.net/?f=%5Csin%20x-2%2B%5Csin%20%5E%7B2%7Dx%3D0)
![\sin ^{2}x+\sin x-2=0](https://tex.z-dn.net/?f=%5Csin%20%5E%7B2%7Dx%2B%5Csin%20x-2%3D0)
Let sin x = u
![u^2+u-2=0](https://tex.z-dn.net/?f=u%5E2%2Bu-2%3D0)
Factor the quadratic equation.
![(u+2)(u-1)=0](https://tex.z-dn.net/?f=%28u%2B2%29%28u-1%29%3D0)
u + 2 = 0, u – 1 = 0
u = –2, u = 1
That is sin x = –2, sin x = 1
sin x can't be smaller than –1 for real solutions. So ignore sin x = –2.
sin x = 1
The value of sin is 1 for 90°.
x = 90°.
Option C is the correct answer.
Answer:
x = StartFraction negative 2 plus or minus StartRoot 2 squared minus 4(4)(negative 1) EndRoot Over 2(4) EndFraction
Step-by-step explanation:
we know that
The formula to solve a quadratic equation of the form
![ax^2+bx+c=0](https://tex.z-dn.net/?f=ax%5E2%2Bbx%2Bc%3D0)
is equal to
![x=\frac{-b(+/-)\sqrt{b^2-4ac} }{2a}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-b%28%2B%2F-%29%5Csqrt%7Bb%5E2-4ac%7D%20%7D%7B2a%7D)
in this problem we have
![4x^2 + 2x - 1=0](https://tex.z-dn.net/?f=4x%5E2%20%2B%202x%20-%201%3D0)
so
![a=4](https://tex.z-dn.net/?f=a%3D4)
![b=2](https://tex.z-dn.net/?f=b%3D2)
![c=-1](https://tex.z-dn.net/?f=c%3D-1)
substitute in the formula
![x=\frac{-2(+/-)\sqrt{2^2-4(4)(-1)} }{2(4)}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-2%28%2B%2F-%29%5Csqrt%7B2%5E2-4%284%29%28-1%29%7D%20%7D%7B2%284%29%7D)
therefore
x = StartFraction negative 2 plus or minus StartRoot 2 squared minus 4(4)(negative 1) EndRoot Over 2(4) EndFraction
That's very interesting. I had never thought about it before.
Let's look through all of the ten possible digits in that place,
and see what we can tell:
-- 0:
A number greater than 10 with a 0 in the units place is a multiple of
either 5 or 10, so it's not a prime number.
-- 1:
A number greater than 10 with a 1 in the units place could be
a prime (11, 31 etc.) but it doesn't have to be (21, 51).
-- 2:
A number greater than 10 with a 2 in the units place has 2 as a factor
(it's an even number), so it's not a prime number.
-- 3:
A number greater than 10 with a 3 in the units place could be
a prime (13, 23 etc.) but it doesn't have to be (33, 63) .
-- 4:
A number greater than 10 with a 4 in the units place is an even
number, and has 2 as a factor, so it's not a prime number.
-- 5:
A number greater than 10 with a 5 in the units place is a multiple
of either 5 or 10, so it's not a prime number.
-- 6:
A number greater than 10 with a 6 in the units place is an even
number, and has 2 as a factor, so it's not a prime number.
-- 7:
A number greater than 10 with a 7 in the units place could be
a prime (17, 37 etc.) but it doesn't have to be (27, 57) .
-- 8:
A number greater than 10 with a 8 in the units place is an even
number, and has 2 as a factor, so it's not a prime number.
-- 9:
A number greater than 10 with a 9 in the units place could be
a prime (19, 29 etc.) but it doesn't have to be (39, 69) .
So a number greater than 10 that IS a prime number COULD have
any of the digits 1, 3, 7, or 9 in its units place.
It CAN't have a 0, 2, 4, 5, 6, or 8 .
The only choice that includes all of the possibilities is 'A' .