Asked by Akingz
                Solve the following equation for values of theta between 0 and 360 degrees
2sec²theta-3tan theta - 1=0
            
        2sec²theta-3tan theta - 1=0
Answers
                    Answered by
            Akingz
            
    Solve the following equation for values of theta between 0 and 360 degrees
8sin²theta + 6cos theta - 9= 0
    
8sin²theta + 6cos theta - 9= 0
                    Answered by
            rithvik
            
    θ
=
2.678
or
2.3562
radians
Explanation:
As
sec
2
θ
=
1
+
tan
2
θ
2
sec
2
θ
+
3
tan
θ
=
1
can be written as
2
+
2
tan
2
θ
+
3
tan
θ
=
1
or
2
tan
θ
+
3
tan
θ
+
1
=
0
i.e.
(
2
tan
θ
+
1
)
(
tan
θ
+
1
)
=
0
Hence either
tan
θ
=
−
1
2
=
tan
(
π
−
0.4636
)
=
tan
2.678
(in radians)
or
tan
θ
=
−
1
=
tan
(
3
π
4
)
=
tan
2.3562
(in radians)
These are results in
[
0
,
π
]
    
=
2.678
or
2.3562
radians
Explanation:
As
sec
2
θ
=
1
+
tan
2
θ
2
sec
2
θ
+
3
tan
θ
=
1
can be written as
2
+
2
tan
2
θ
+
3
tan
θ
=
1
or
2
tan
θ
+
3
tan
θ
+
1
=
0
i.e.
(
2
tan
θ
+
1
)
(
tan
θ
+
1
)
=
0
Hence either
tan
θ
=
−
1
2
=
tan
(
π
−
0.4636
)
=
tan
2.678
(in radians)
or
tan
θ
=
−
1
=
tan
(
3
π
4
)
=
tan
2.3562
(in radians)
These are results in
[
0
,
π
]
                    Answered by
            rithvik
            
    for the second one, Step 1. Find the value of :
Given,
 
 
 
Divide the above equation by 5:
 
 
 
 
 
 
Step 2. Let
 
 
and
 
 
.(Consider)
 
 
 
 
 
are the solutions of the given equation.
or
 
 
Step 3. Take “sin” on both sides, we get
 
 
 
    
Given,
Divide the above equation by 5:
Step 2. Let
and
.(Consider)
are the solutions of the given equation.
or
Step 3. Take “sin” on both sides, we get
                    Answered by
            oobleck
            
    2sec^2θ-3tanθ - 1 = 0
2(1+tan^2θ)-3tanθ-1 = 0
2tan^2θ - 3tanθ + 1 = 0
(2tanθ-1)(tanθ-1) = 0
tanθ = -1/2
tanθ = 1
now finish it off
same idea.
8sin^2θ + 6cosθ - 9= 0
8cos^2θ - 6cosθ + 1 = 0
(4cosθ - 1)(2cosθ - 1) = 0
cosθ = 1/4
cosθ = 1/2
and so on
    
2(1+tan^2θ)-3tanθ-1 = 0
2tan^2θ - 3tanθ + 1 = 0
(2tanθ-1)(tanθ-1) = 0
tanθ = -1/2
tanθ = 1
now finish it off
same idea.
8sin^2θ + 6cosθ - 9= 0
8cos^2θ - 6cosθ + 1 = 0
(4cosθ - 1)(2cosθ - 1) = 0
cosθ = 1/4
cosθ = 1/2
and so on
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