What is the general solution for cot(2θ)+csc(2θ)-7=0 ?

The general solution for cot(2θ)+csc(2θ)-7=0 is θ=(0.28379…)/2+pin

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cot(2θ)+csc(2θ)-7=0

cot (2 θ) + csc (2 θ) - 7 = 0

θ = \frac{0.28379 \dots}{2} + πn

Solution steps

cot (2 θ) + csc (2 θ) - 7 = 0

Express with sin, cos

\frac{cos ( 2 θ )}{sin ( 2 θ )} + \frac{1}{sin ( 2 θ )} - 7 = 0

Simplify

\frac{cos ( 2 θ )}{sin ( 2 θ )} + \frac{1}{sin ( 2 θ )} - 7 : \frac{cos ( 2 θ ) + 1 - 7 sin ( 2 θ )}{sin ( 2 θ )}

\frac{cos ( 2 θ ) + 1 - 7 sin ( 2 θ )}{sin ( 2 θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

cos (2 θ) + 1 - 7 sin (2 θ) = 0

Add

7 sin (2 θ)

to both sides

cos (2 θ) + 1 = 7 sin (2 θ)

Square both sides

(cos (2 θ) + 1)^{2} = (7 sin (2 θ))^{2}

Subtract

(7 sin (2 θ))^{2}

from both sides

(cos (2 θ) + 1)^{2} - 49 sin^{2} (2 θ) = 0

Rewrite using trig identities

- 48 + 2 cos (2 θ) + 50 cos^{2} (2 θ) = 0

Solve by substitution

cos (2 θ) = \frac{24}{25}, cos (2 θ) = - 1

cos (2 θ) = \frac{24}{25} : θ = \frac{arccos ( \frac{24}{25} )}{2} + πn, θ = π - \frac{arccos ( \frac{24}{25} )}{2} + πn

cos (2 θ) = - 1 : θ = \frac{π}{2} + πn

Combine all the solutions

θ = \frac{arccos ( \frac{24}{25} )}{2} + πn, θ = π - \frac{arccos ( \frac{24}{25} )}{2} + πn, θ = \frac{π}{2} + πn

Verify solutions by plugging them into the original equation

θ = \frac{arccos ( \frac{24}{25} )}{2} + πn

Show solutions in decimal form

θ = \frac{0.28379 \dots}{2} + πn

sec (x) sin (x) + cos (x) = sec (x)

0 = 2 + 2 sin (θ)

x, 2 cos (x) + 3 = 0

4 cos (x) - 5 = - 8

sin (x) + 4 csc (x) + 5 = 0, 0 \leq x \leq 2 π

What is the general solution for cot(2θ)+csc(2θ)-7=0 ?
The general solution for cot(2θ)+csc(2θ)-7=0 is θ=(0.28379…)/2+pin