tan x = √3
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tan x = √3

sec x = 2


cot x = – √3


cosec x = – 2


Find the general solution for each of the following equations:
cos 4x = cos 2x

cos 3x + cos x – cos 2x = 0

sin 2x + cos x = 0
It is given that
sin 2x + cos x = 0
We can write it as
2 sin x cos x + cos x = 0
cos x (2 sin x + 1) = 0
cos x = 0 or 2 sin x + 1 = 0
Let cos x = 0

sec2 2x = 1 – tan 2x
It is given that
sec2 2x = 1 – tan 2x
We can write it as
1 + tan2 2x = 1 – tan 2x
tan2 2x + tan 2x = 0
Taking common terms
tan 2x (tan 2x + 1) = 0
Here
tan 2x = 0 or tan 2x + 1 = 0
If tan 2x = 0
tan 2x = tan 0
We get
2x = nπ + 0, where n ∈ Z
x = nπ/2, where n ∈ Z
tan 2x + 1 = 0
We can write it as
tan 2x = – 1
So we get

Here
2x = nπ + 3π/4, where n ∈ Z
x = nπ/2 + 3π/8, where n ∈ Z
Hence, the general solution is nπ/2 or nπ/2 + 3π/8, n ∈ Z.
sin x + sin 3x + sin 5x = 0
It is given that
sin x + sin 3x + sin 5x = 0
We can write it as
(sin x + sin 5x) + sin 3x = 0
Using the formula

By further calculation
2 sin 3x cos (-2x) + sin 3x = 0
It can be written as
2 sin 3x cos 2x + sin 3x = 0
By taking out the common terms
sin 3x (2 cos 2x + 1) = 0
Here
sin 3x = 0 or 2 cos 2x + 1 = 0
If sin 3x = 0
3x = nπ, where n ∈ Z
We get
x = nπ/3, where n ∈ Z
If 2 cos 2x + 1 = 0
cos 2x = – 1/2
By further simplification
= – cos π/3
= cos (π – π/3)
So we get
cos 2x = cos 2π/3
Here
