What is the general solution for sqrt(2)sin(x)sec(x)=2sin(x) ?

The general solution for sqrt(2)sin(x)sec(x)=2sin(x) is x=2pin,x=pi+2pin,x= pi/4+2pin,x=(7pi)/4+2pin

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sqrt(2)sin(x)sec(x)=2sin(x)

Solution

2 sin (x) sec (x) = 2 sin (x)

Solution

x = 2 πn, x = π + 2 πn, x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n

Solution steps

2 sin (x) sec (x) = 2 sin (x)

Subtract

2 sin (x)

from both sides

2 sin (x) sec (x) - 2 sin (x) = 0

Factor

2 sin (x) sec (x) - 2 sin (x) : sin (x) (2 sec (x) - 2)

2 sin (x) sec (x) - 2 sin (x)

Factor out common term

sin (x)

= sin (x) (2 sec (x) - 2)

sin (x) (2 sec (x) - 2) = 0

Solving each part separately

sin (x) = 0 or 2 sec (x) - 2 = 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

2 sec (x) - 2 = 0 : x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

2 sec (x) - 2 = 0

Move

2

to the right side

2 sec (x) - 2 = 0

Add

2

to both sides

2 sec (x) - 2 + 2 = 0 + 2

Simplify

2 sec (x) = 2

2 sec (x) = 2

Divide both sides by

2

2 sec (x) = 2

Divide both sides by

2

\frac{2 sec ( x )}{2} = \frac{2}{2}

Simplify

\frac{2 sec ( x )}{2} = \frac{2}{2}

Simplify

\frac{2 sec ( x )}{2} : sec (x)

\frac{2 sec ( x )}{2}

Cancel the common factor:

2

= sec (x)

Simplify

\frac{2}{2} : 2

\frac{2}{2}

Apply radical rule:

2 = 2^{\frac{1}{2}}

= \frac{2}{2 ^{\frac{1}{2}}}

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

\frac{2 ^{1}}{2 ^{\frac{1}{2}}} = 2^{1 - \frac{1}{2}}

= 2^{1 - \frac{1}{2}}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= 2^{\frac{1}{2}}

Apply radical rule:

2^{\frac{1}{2}} = 2

= 2

sec (x) = 2

sec (x) = 2

sec (x) = 2

General solutions for

sec (x) = 2

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

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sqrt(3)-2sin(x)=0

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-9cos^2(θ)+14sin(θ)+14=0

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cos(x)tan(x)=-7tan(x)

cos (x) tan (x) = - 7 tan (x)

12sin^2(θ)-9=0

12 sin^{2} (θ) - 9 = 0

Frequently Asked Questions (FAQ)

What is the general solution for sqrt(2)sin(x)sec(x)=2sin(x) ?
The general solution for sqrt(2)sin(x)sec(x)=2sin(x) is x=2pin,x=pi+2pin,x= pi/4+2pin,x=(7pi)/4+2pin