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

The general solution for 2-sin(x)tan(x)=cos(x) is x= pi/3+2pin,x=(5pi)/3+2pin

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Popular >

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

Solution

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

Solution

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Degrees

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

cos (x)

from both sides

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

Express with sin, cos

2 - cos (x) - sin (x) tan (x)

Use the basic trigonometric identity:

tan (x) = \frac{sin ( x )}{cos ( x )}

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

Simplify

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

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

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

sin (x) \frac{sin ( x )}{cos ( x )}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

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

sin (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (x) sin (x) = sin^{1 + 1} (x)

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

Convert element to fraction:

2 = \frac{2 cos ( x )}{cos ( x )}, cos (x) = \frac{cos ( x ) cos ( x )}{cos ( x )}

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

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

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

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

cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (x) cos (x) = cos^{1 + 1} (x)

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

- cos^{2} (x) - sin^{2} (x) = - 1

= 2 cos (x) - 1

2 cos (x) - 1 = 0

Move

1

to the right side

2 cos (x) - 1 = 0

Add

1

to both sides

2 cos (x) - 1 + 1 = 0 + 1

Simplify

2 cos (x) = 1

2 cos (x) = 1

Divide both sides by

2

2 cos (x) = 1

Divide both sides by

2

\frac{2 cos ( x )}{2} = \frac{1}{2}

Simplify

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{2}

General solutions for

cos (x) = \frac{1}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Popular Examples

tan(2x-n/4)=-1,0<x<= 360

tan (2 x - \frac{n}{4}) = - 1, 0^{\circ} < x \leq 36 0^{\circ}

2sin^2(θ)=cos^2(θ)

2 sin^{2} (θ) = cos^{2} (θ)

-cos(x)=-sin^2(x)-1,(0,2pi)

- cos (x) = - sin^{2} (x) - 1, (0, 2 π)

solvefor x,y+arctan(y/x)=0 solve for

x, y + arctan (\frac{y}{x}) = 0

tan(θ)=-1,0<= θ<= 2pi

tan (θ) = - 1, 0 \leq θ \leq 2 π

Frequently Asked Questions (FAQ)

What is the general solution for 2-sin(x)tan(x)=cos(x) ?
The general solution for 2-sin(x)tan(x)=cos(x) is x= pi/3+2pin,x=(5pi)/3+2pin