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

The general solution for sin^2(x)-tan(x)cos^2(x)=0 is x=2pin,x=pi+2pin,x= pi/4+pin

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

Solution

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

Solution

x = 2 πn, x = π + 2 πn, x = \frac{π}{4} + πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

= sin (x) cos (x)

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

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

Factor

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

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

Apply exponent rule:

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

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

= sin (x) sin (x) - sin (x) cos (x)

Factor out common term

sin (x)

= sin (x) (sin (x) - cos (x))

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

Solving each part separately

sin (x) = 0 or sin (x) - cos (x) = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) = 0

General solutions for

sin (x) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

sin (x) - cos (x) = 0 : x = \frac{π}{4} + πn

sin (x) - cos (x) = 0

Rewrite using trig identities

sin (x) - cos (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

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

Simplify

\frac{sin ( x )}{cos ( x )} - 1 = 0

Use the basic trigonometric identity:

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

tan (x) - 1 = 0

tan (x) - 1 = 0

Move

1

to the right side

tan (x) - 1 = 0

Add

1

to both sides

tan (x) - 1 + 1 = 0 + 1

Simplify

tan (x) = 1

tan (x) = 1

General solutions for

tan (x) = 1

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{π}{4} + πn

Popular Examples

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

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

sec^2(x)=2tan(x)

sec^{2} (x) = 2 tan (x)

4sin^2(x)-8cos(x)+1=0

4 sin^{2} (x) - 8 cos (x) + 1 = 0

sec^2(θ)=4

sec^{2} (θ) = 4

sin(a)= 5/13

sin (a) = \frac{5}{13}

Frequently Asked Questions (FAQ)

What is the general solution for sin^2(x)-tan(x)cos^2(x)=0 ?
The general solution for sin^2(x)-tan(x)cos^2(x)=0 is x=2pin,x=pi+2pin,x= pi/4+pin