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شائع علم المثلثات >

tan(-x)=tan(2x+1)

الحلّ

tan (- x) = tan (2 x + 1)

الحلّ

x = \frac{2 πn}{3} - \frac{1}{3}, x = \frac{π}{3} + \frac{2 πn}{3} - \frac{1}{3}

درجات

x = - 19.09859 \dots^{\circ} + 12 0^{\circ} n, x = 40.90140 \dots^{\circ} + 12 0^{\circ} n

خطوات الحلّ

tan (- x) = tan (2 x + 1)

Rewrite using trig identities

tan (- x) = tan (2 x + 1)

- tan (x) = tan (2 x + 1)

- tan (x) = tan (2 x + 1)

من الطرفين

tan (2 x + 1)

اطرح

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

sin, cos :

عبّر بواسطة

- tan (1 + 2 x) - tan (x)

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

:Use the basic trigonometric identity

= - \frac{sin ( 1 + 2 x )}{cos ( 1 + 2 x )} - tan (x)

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

:Use the basic trigonometric identity

= - \frac{sin ( 1 + 2 x )}{cos ( 1 + 2 x )} - \frac{sin ( x )}{cos ( x )}

- \frac{sin ( 1 + 2 x )}{cos ( 1 + 2 x )} - \frac{sin ( x )}{cos ( x )}

بسّط:

\frac{- sin ( 1 + 2 x ) cos ( x ) - sin ( x ) cos ( 2 x + 1 )}{cos ( 2 x + 1 ) cos ( x )}

- \frac{sin ( 1 + 2 x )}{cos ( 1 + 2 x )} - \frac{sin ( x )}{cos ( x )}

cos (1 + 2 x), cos (x)

المضاعف المشترك الأصغر لـ:

cos (2 x + 1) cos (x)

cos (1 + 2 x), cos (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (1 + 2 x)

cos (x)

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

اكتب مجددًا الكسور بحيث يكون المقام مشترك

cos (2 x + 1) cos (x)

اضرب كل بسط ومقام بتعبير الذي يؤدّي إلى مقام مشترك

For

\frac{sin ( 1 + 2 x )}{cos ( 1 + 2 x )} :

multiply the denominator and numerator by

cos (x)

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

For

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

multiply the denominator and numerator by

cos (2 x + 1)

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

= - \frac{sin ( 1 + 2 x ) cos ( x )}{cos ( 1 + 2 x ) cos ( x )} - \frac{sin ( x ) cos ( 2 x + 1 )}{cos ( x ) cos ( 2 x + 1 )}

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

:بما أنّ المقامات متساوية، اجمع البسوط

= \frac{- sin ( 1 + 2 x ) cos ( x ) - sin ( x ) cos ( 2 x + 1 )}{cos ( 2 x + 1 ) cos ( x )}

= \frac{- sin ( 1 + 2 x ) cos ( x ) - sin ( x ) cos ( 2 x + 1 )}{cos ( 2 x + 1 ) cos ( x )}

\frac{- cos ( 1 + 2 x ) sin ( x ) - cos ( x ) sin ( 1 + 2 x )}{cos ( 1 + 2 x ) cos ( x )} = 0

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

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

Rewrite using trig identities

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

sin (s) cos (t) + cos (s) sin (t) = sin (s + t)

:فعّل متطابقة الجمع لزوايا

- sin (s) cos (t) - cos (s) sin (t) = - sin (s + t)

= - sin (1 + 2 x + x)

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

- 1

اقسم الطرفين على

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

- 1

اقسم الطرفين على

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

بسّط

sin (1 + 2 x + x) = 0

sin (1 + 2 x + x) = 0

sin (1 + 2 x + x) = 0 :

حلول عامّة لـ

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}

1 + 2 x + x = 0 + 2 πn, 1 + 2 x + x = π + 2 πn

1 + 2 x + x = 0 + 2 πn, 1 + 2 x + x = π + 2 πn

1 + 2 x + x = 0 + 2 πn

حلّ:

x = \frac{2 πn}{3} - \frac{1}{3}

1 + 2 x + x = 0 + 2 πn

2 x + x = 3 x :

اجمع العناصر المتشابهة

1 + 3 x = 0 + 2 πn

0 + 2 πn = 2 πn

1 + 3 x = 2 πn

انقل

1

إلى الجانب الأيمن

1 + 3 x = 2 πn

من الطرفين

1

اطرح

1 + 3 x - 1 = 2 πn - 1

بسّط

3 x = 2 πn - 1

3 x = 2 πn - 1

3

اقسم الطرفين على

3 x = 2 πn - 1

3

اقسم الطرفين على

\frac{3 x}{3} = \frac{2 πn}{3} - \frac{1}{3}

بسّط

x = \frac{2 πn}{3} - \frac{1}{3}

x = \frac{2 πn}{3} - \frac{1}{3}

1 + 2 x + x = π + 2 πn

حلّ:

x = \frac{π}{3} + \frac{2 πn}{3} - \frac{1}{3}

1 + 2 x + x = π + 2 πn

2 x + x = 3 x :

اجمع العناصر المتشابهة

1 + 3 x = π + 2 πn

انقل

1

إلى الجانب الأيمن

1 + 3 x = π + 2 πn

من الطرفين

1

اطرح

1 + 3 x - 1 = π + 2 πn - 1

بسّط

3 x = π + 2 πn - 1

3 x = π + 2 πn - 1

3

اقسم الطرفين على

3 x = π + 2 πn - 1

3

اقسم الطرفين على

\frac{3 x}{3} = \frac{π}{3} + \frac{2 πn}{3} - \frac{1}{3}

بسّط

x = \frac{π}{3} + \frac{2 πn}{3} - \frac{1}{3}

x = \frac{π}{3} + \frac{2 πn}{3} - \frac{1}{3}

x = \frac{2 πn}{3} - \frac{1}{3}, x = \frac{π}{3} + \frac{2 πn}{3} - \frac{1}{3}

رسم

أمثلة شائعة

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

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

sin(6x)cos(9x)-cos(6x)sin(9x)=-0.4

sin (6 x) cos (9 x) - cos (6 x) sin (9 x) = - 0.4

sin^2(x)=2tan^2(x)

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

cos^3(x)=cos(2x+x)

cos^{3} (x) = cos (2 x + x)

solvefor x,(d^2-3d+2)y=sin(e^x)

so l v e f or x, (d^{2} - 3 d + 2) y = sin (e^{x})