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Find the break even points of the expression $\frac{\left(\sin\left(x\right)+\cos\left(x\right)\right)^2}{1+2\sin\left(x\right)\cos\left(x\right)}$

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Final answer to the problem

$1$
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Step-by-step Solution

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Expand the expression $\left(\sin\left(x\right)+\cos\left(x\right)\right)^2$ using the square of a binomial: $(a+b)^2=a^2+2ab+b^2$

$\frac{\sin\left(x\right)^{2}+2\sin\left(x\right)\cos\left(x\right)+\cos\left(x\right)^{2}}{1+2\sin\left(x\right)\cos\left(x\right)}$

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$\frac{\sin\left(x\right)^{2}+2\sin\left(x\right)\cos\left(x\right)+\cos\left(x\right)^{2}}{1+2\sin\left(x\right)\cos\left(x\right)}$

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Learn how to solve classify algebraic expressions problems step by step online. Find the break even points of the expression ((sin(x)+cos(x))^2)/(1+2sin(x)cos(x)). Expand the expression \left(\sin\left(x\right)+\cos\left(x\right)\right)^2 using the square of a binomial: (a+b)^2=a^2+2ab+b^2. Applying the pythagorean identity: \sin^2\left(\theta\right)+\cos^2\left(\theta\right)=1. Simplify the fraction \frac{1+2\sin\left(x\right)\cos\left(x\right)}{1+2\sin\left(x\right)\cos\left(x\right)} by 1+2\sin\left(x\right)\cos\left(x\right).

Final answer to the problem

$1$

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7
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9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

How to improve your answer:

Main Topic: Classify algebraic expressions

An algebraic expression can be classified as a monomial, binomial, trinomial or polynomial, depending on the number of terms.

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