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Solve the trigonometric equation $\frac{2\cos\left(x\right)^2+\cos\left(x\right)-3}{\sin\left(x\right)^2}=\frac{2\cos\left(x\right)-3}{\cos\left(x\right)+1}$

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Solution

Final answer to the problem

$x=\frac{1}{2}\pi+2\pi n,\:x=\frac{3}{2}\pi+2\pi n,\:x=0+2\pi n,\:x=\pi+2\pi n\:,\:\:n\in\Z$
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Applying the trigonometric identity: $\sin\left(\theta \right)^2 = 1-\cos\left(\theta \right)^2$

$\frac{2\cos\left(x\right)^2+\cos\left(x\right)-3}{1-\cos\left(x\right)^2}=\frac{2\cos\left(x\right)-3}{\cos\left(x\right)+1}$
Why is 1 - cos(x)^2 = sin(x)^2 ?

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$\frac{2\cos\left(x\right)^2+\cos\left(x\right)-3}{1-\cos\left(x\right)^2}=\frac{2\cos\left(x\right)-3}{\cos\left(x\right)+1}$

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Learn how to solve problems step by step online. Solve the trigonometric equation (2cos(x)^2+cos(x)+-3)/(sin(x)^2)=(2cos(x)-3)/(cos(x)+1). Applying the trigonometric identity: \sin\left(\theta \right)^2 = 1-\cos\left(\theta \right)^2. Apply fraction cross-multiplication. Multiplying polynomials 2\cos\left(x\right)^2+\cos\left(x\right)-3 and \cos\left(x\right)+1. Multiply the single term \cos\left(x\right) by each term of the polynomial \left(2\cos\left(x\right)^2+\cos\left(x\right)-3\right).

Final answer to the problem

$x=\frac{1}{2}\pi+2\pi n,\:x=\frac{3}{2}\pi+2\pi n,\:x=0+2\pi n,\:x=\pi+2\pi n\:,\:\:n\in\Z$

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Function Plot

Plotting: $\frac{2\cos\left(x\right)^2+\cos\left(x\right)-3}{\sin\left(x\right)^2}+\frac{-2\cos\left(x\right)+3}{\cos\left(x\right)+1}$

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

◻√◻
◻√
∞
e
π
ln
log
log◻
lim
d/dx
D□x
∫
∫◻
|◻|
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

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