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Answer:
-1, 4+i, 4-i
Step-by-step explanation:
x^4- 6x^3 + 2x^2 + 26x + 17
Using the rational root theorem
we see if 1, -1, -17 or 17 are roots
Check and see if 1 is a root
1^4- 6(1^3) + 2(1^2) + 26(1) + 17=0
1-6+2+26+17 does not equal 0 1 is not a root
-1
1^4- 6(-1^3) + 2(1^2) + 26(-1) + 17=0
1 +6 +2 -26+17 = 0
-1 is a root
Factor out (x+1)
(x+1) ( x^3-7x^2+9x+17)
Using the rational root theorem again on x^3-7x^2+9x+17
Checking -1
-1 -7 -9 +17=0
-1 is a root
(x+1) (x+1) (x^2-8x+17)
Using the quadratic on the last
8 ±sqrt(8^2 - 4(1)17)
--------------------------------
2
gives imaginary roots
4±i
<u>Given</u>:
Given that we need to prove the identity 
<u>Proof</u>:
Step 1: Factor out the common term sin x, we get;
Step 2: Using the identity 
Step 3: Reciprocating sec x, we get;
Step 4: Splitting the denominator, we have;
Simplifying, we get;
Thus, the identity is proved.
Answer:
first one yes
Step-by-step explanation:
second one its
39x+10=180
