Option C:
x = 90°
Solution:
Given equation:

<u>To find the degree:</u>

Subtract 1 + cos²x from both sides.

Using the trigonometric identity:




Let sin x = u

Factor the quadratic equation.

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

is equal to

in this problem we have

so



substitute in the formula

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' .