What is the general solution for 2cos(x)-tan(x)=0 ?

The general solution for 2cos(x)-tan(x)=0 is x=0.89590…+2pin,x=pi-0.89590…+2pin

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2cos(x)-tan(x)=0

2 cos (x) - tan (x) = 0

x = 0.89590 \dots + 2 πn, x = π - 0.89590 \dots + 2 πn

Solution steps

2 cos (x) - tan (x) = 0

Express with sin, cos

2 cos (x) - \frac{sin ( x )}{cos ( x )} = 0

Simplify

2 cos (x) - \frac{sin ( x )}{cos ( x )} : \frac{2 cos ^{2} ( x ) - sin ( x )}{cos ( x )}

\frac{2 cos ^{2} ( x ) - sin ( x )}{cos ( x )} = 0

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

2 cos^{2} (x) - sin (x) = 0

Add

sin (x)

to both sides

2 cos^{2} (x) = sin (x)

Square both sides

(2 cos^{2} (x))^{2} = sin^{2} (x)

Subtract

sin^{2} (x)

from both sides

4 cos^{4} (x) - sin^{2} (x) = 0

Factor

4 cos^{4} (x) - sin^{2} (x) : (2 cos^{2} (x) + sin (x)) (2 cos^{2} (x) - sin (x))

(2 cos^{2} (x) + sin (x)) (2 cos^{2} (x) - sin (x)) = 0

Solving each part separately

2 cos^{2} (x) + sin (x) = 0 or 2 cos^{2} (x) - sin (x) = 0

2 cos^{2} (x) + sin (x) = 0 : x = arcsin (- \frac{- 1 + 17}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 17}{4}) + 2 πn

2 cos^{2} (x) - sin (x) = 0 : x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

Combine all the solutions

x = arcsin (- \frac{- 1 + 17}{4}) + 2 πn, x = π + arcsin (\frac{- 1 + 17}{4}) + 2 πn, x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

Verify solutions by plugging them into the original equation

x = arcsin (\frac{17 - 1}{4}) + 2 πn, x = π - arcsin (\frac{17 - 1}{4}) + 2 πn

Show solutions in decimal form

x = 0.89590 \dots + 2 πn, x = π - 0.89590 \dots + 2 πn

0 = 3 sin (x)

sin (\frac{θ}{2}) = 0.4

arccos (x) = \frac{79}{9 70}

sin (2 x) sin (x) - cos (2 x) cos (x) = \frac{2}{2}

50 sec^{2} (5 x) tan (5 x) = 25 + tan^{2} (5 x)

What is the general solution for 2cos(x)-tan(x)=0 ?
The general solution for 2cos(x)-tan(x)=0 is x=0.89590…+2pin,x=pi-0.89590…+2pin