What is the general solution for tan(2x-5)=sin^2(30)+sin^2(60) ?

The general solution for tan(2x-5)=sin^2(30)+sin^2(60) is x= 5/2+22.5+(180n)/2

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tan(2x-5)=sin^2(30)+sin^2(60)

Solution

tan (2 x - 5) = sin^{2} (3 0^{\circ}) + sin^{2} (6 0^{\circ})

Solution

x = \frac{5}{2} + 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

Radians

x = \frac{5}{2} + \frac{π}{8} + \frac{π}{2} n

Solution steps

tan (2 x - 5) = sin^{2} (3 0^{\circ}) + sin^{2} (6 0^{\circ})

sin (3 0^{\circ}) = \frac{1}{2}

sin (3 0^{\circ})

Use the following trivial identity:

sin (3 0^{\circ}) = \frac{1}{2}

sin (3 0^{\circ})

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

= \frac{1}{2}

sin (6 0^{\circ}) = \frac{3}{2}

sin (6 0^{\circ})

Use the following trivial identity:

sin (6 0^{\circ}) = \frac{3}{2}

sin (6 0^{\circ})

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{3}{2}

= \frac{3}{2}

tan (2 x - 5) = (\frac{1}{2})^{2} + (\frac{3}{2})^{2}

Simplify

(\frac{1}{2})^{2} + (\frac{3}{2})^{2} : 1

(\frac{1}{2})^{2} + (\frac{3}{2})^{2}

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

(\frac{1}{2})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply rule

1^{a} = 1

1^{2} = 1

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

(\frac{3}{2})^{2} = \frac{3}{2 ^{2}}

(\frac{3}{2})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{( 3 ) ^{2}}{2 ^{2}}

(3)^{2} : 3

Apply radical rule:

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

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= 3^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 3

= \frac{3}{2 ^{2}}

= \frac{1}{2 ^{2}} + \frac{3}{2 ^{2}}

Apply rule

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

= \frac{1 + 3}{2 ^{2}}

Add the numbers:

1 + 3 = 4

= \frac{4}{2 ^{2}}

Factor

4 : 2^{2}

Factor

4 = 2^{2}

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

Cancel the common factor:

2^{2}

= 1

General solutions for

tan (2 x - 5) = 1

tan (x)

periodicity table with

18 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

2 x - 5 = 4 5^{\circ} + 18 0^{\circ} n

2 x - 5 = 4 5^{\circ} + 18 0^{\circ} n

Solve

2 x - 5 = 4 5^{\circ} + 18 0^{\circ} n : x = \frac{5}{2} + 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

2 x - 5 = 4 5^{\circ} + 18 0^{\circ} n

Move

5

to the right side

2 x - 5 = 4 5^{\circ} + 18 0^{\circ} n

Add

5

to both sides

2 x - 5 + 5 = 4 5^{\circ} + 18 0^{\circ} n + 5

Simplify

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

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

Divide both sides by

2

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

Divide both sides by

2

\frac{2 x}{2} = \frac{4 5 ^{\circ}}{2} + \frac{18 0 ^{\circ} n}{2} + \frac{5}{2}

Simplify

\frac{2 x}{2} = \frac{4 5 ^{\circ}}{2} + \frac{18 0 ^{\circ} n}{2} + \frac{5}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{4 5 ^{\circ}}{2} + \frac{18 0 ^{\circ} n}{2} + \frac{5}{2} : \frac{5}{2} + 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

\frac{4 5 ^{\circ}}{2} + \frac{18 0 ^{\circ} n}{2} + \frac{5}{2}

Group like terms

= \frac{5}{2} + \frac{18 0 ^{\circ} n}{2} + \frac{4 5 ^{\circ}}{2}

\frac{4 5 ^{\circ}}{2} = 22. 5^{\circ}

\frac{4 5 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{18 0 ^{\circ}}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= 22. 5^{\circ}

= \frac{5}{2} + \frac{18 0 ^{\circ} n}{2} + 22. 5^{\circ}

Group like terms

= \frac{5}{2} + 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

x = \frac{5}{2} + 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

x = \frac{5}{2} + 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

x = \frac{5}{2} + 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

x = \frac{5}{2} + 22. 5^{\circ} + \frac{18 0 ^{\circ} n}{2}

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Frequently Asked Questions (FAQ)

What is the general solution for tan(2x-5)=sin^2(30)+sin^2(60) ?
The general solution for tan(2x-5)=sin^2(30)+sin^2(60) is x= 5/2+22.5+(180n)/2