2cos^2 x + 3sin x = 0
2(1 - sin^2 x) + 3sin x = 0
2 - 2sin^2 x + 3sin x = 0
2sin^2 x - 3sin x - 2 = 0
Let sin x = m, then
2m^2 - 3m - 2 = 0
2m^2 + m - 4m - 2 = 0
m(2m + 1) - 2(2m + 1) = 0
(m - 2)(2m + 1) = 0
m = 2 or m = -1/2
Now, sin x = -1/2
Therefore, x = 1/6(12nπ - π) and 1/6(12nπ + 7π)
Answer:Classifying Polynomials. Polynomials can be classified two different ways - by the number of terms and by their degree. A monomial has just one term. For example, 4x2 .Remember that a term contains both the variable(s) and its coefficient (the number in front of it.)
Step-by-step explanation:Hope this helps. Please name me brainliest
Protons my dude.............
No, they are both incorrect by my understanding of your question use the Pythagorean theorem a^2 + b^2 =c^, therefore, 7^2 +x^2 = 13^2 or 49+x=169 then solve normally 169-49=120 √120 =5
The answer is 5
