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شائع علم المثلثات >

2tan(x)+sec(x)=0,0<= x<2pi

الحلّ

2 tan (x) + sec (x) = 0, 0 \leq x < 2 π

الحلّ

x = \frac{7 π}{6}, x = \frac{11 π}{6}

+1

درجات

x = 21 0^{\circ}, x = 33 0^{\circ}

خطوات الحلّ

2 tan (x) + sec (x) = 0, 0 \leq x < 2 π

sin, cos :

عبّر بواسطة

sec (x) + 2 tan (x)

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

:Use the basic trigonometric identity

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

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

:Use the basic trigonometric identity

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

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

بسّط:

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

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

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

اضرب بـ:

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

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

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

:اضرب كسور

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

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

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

فعّل القانون

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

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

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

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

1 + 2 sin (x) = 0

انقل

1

إلى الجانب الأيمن

1 + 2 sin (x) = 0

من الطرفين

1

اطرح

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

بسّط

2 sin (x) = - 1

2 sin (x) = - 1

2

اقسم الطرفين على

2 sin (x) = - 1

2

اقسم الطرفين على

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

بسّط

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

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

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

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}

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

0 \leq x < 2 π :

حلول للمدى

x = \frac{7 π}{6}, x = \frac{11 π}{6}

رسم

أمثلة شائعة

1-sin(x)=sqrt(3)cos(x)

1 - sin (x) = 3 cos (x)

cos(x)-sec(x)=0

cos (x) - sec (x) = 0

2+7cos(x)=4sin^2(x)

2 + 7 cos (x) = 4 sin^{2} (x)

4 arccos (x) = π

2sin^2(x)+2=7cos(x)

2 sin^{2} (x) + 2 = 7 cos (x)