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شائع علم المثلثات >

4cos(x)-2=2tan(x)cot(x)-sec(x)

الحلّ

4 cos (x) - 2 = 2 tan (x) cot (x) - sec (x)

الحلّ

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

درجات

x = 6 0^{\circ} + 36 0^{\circ} n, x = 30 0^{\circ} + 36 0^{\circ} n

خطوات الحلّ

4 cos (x) - 2 = 2 tan (x) cot (x) - sec (x)

من الطرفين

2 tan (x) cot (x) - sec (x)

اطرح

4 cos (x) - 2 - 2 tan (x) cot (x) + sec (x) = 0

sin, cos :

عبّر بواسطة

- 2 + sec (x) + 4 cos (x) - 2 cot (x) tan (x)

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

:Use the basic trigonometric identity

= - 2 + \frac{1}{cos ( x )} + 4 cos (x) - 2 cot (x) tan (x)

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

:Use the basic trigonometric identity

= - 2 + \frac{1}{cos ( x )} + 4 cos (x) - 2 \cdot \frac{cos ( x )}{sin ( x )} tan (x)

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

:Use the basic trigonometric identity

= - 2 + \frac{1}{cos ( x )} + 4 cos (x) - 2 \cdot \frac{cos ( x )}{sin ( x )} \cdot \frac{sin ( x )}{cos ( x )}

- 2 + \frac{1}{cos ( x )} + 4 cos (x) - 2 \cdot \frac{cos ( x )}{sin ( x )} \cdot \frac{sin ( x )}{cos ( x )}

بسّط:

\frac{1 + 4 cos ^{2} ( x ) - 4 cos ( x )}{cos ( x )}

- 2 + \frac{1}{cos ( x )} + 4 cos (x) - 2 \cdot \frac{cos ( x )}{sin ( x )} \cdot \frac{sin ( x )}{cos ( x )}

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

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

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

:اضرب كسور

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

cos (x) :

إلغ العوامل المشتركة

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

sin (x) :

إلغ العوامل المشتركة

= 2

= - 2 + \frac{1}{cos ( x )} + 4 cos (x) - 2

جمّع التعابير المتشابهة

= \frac{1}{cos ( x )} + 4 cos (x) - 2 - 2

- 2 - 2 = - 4 :

اطرح الأعداد

= \frac{1}{cos ( x )} + 4 cos (x) - 4

4 cos (x) = \frac{4 cos ( x ) cos ( x )}{cos ( x )}, 4 = \frac{4 cos ( x )}{cos ( x )}

:حوّل الأعداد لكسور

= \frac{1}{cos ( x )} + \frac{4 cos ( x ) cos ( x )}{cos ( x )} - \frac{4 cos ( x )}{cos ( x )}

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

:بما أنّ المقامات متساوية، اجمع البسوط

= \frac{1 + 4 cos ( x ) cos ( x ) - 4 cos ( x )}{cos ( x )}

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

1 + 4 cos (x) cos (x) - 4 cos (x)

4 cos (x) cos (x) = 4 cos^{2} (x)

4 cos (x) cos (x)

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

:فعّل قانون القوى

cos (x) cos (x) = cos^{1 + 1} (x)

= 4 cos^{1 + 1} (x)

1 + 1 = 2 :

اجمع الأعداد

= 4 cos^{2} (x)

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

= \frac{1 + 4 cos ^{2} ( x ) - 4 cos ( x )}{cos ( x )}

= \frac{1 + 4 cos ^{2} ( x ) - 4 cos ( x )}{cos ( x )}

\frac{1 - 4 cos ( x ) + 4 cos ^{2} ( x )}{cos ( x )} = 0

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

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

بالاستعانة بطريقة التعويض

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

cos (x) = u :

على افتراض أنّ

1 - 4 u + 4 u^{2} = 0

1 - 4 u + 4 u^{2} = 0 : u = \frac{1}{2}

1 - 4 u + 4 u^{2} = 0

a x^{2} + b x + c = 0

اكتب بالصورة الاعتياديّة

4 u^{2} - 4 u + 1 = 0

حلّ بالاستعانة بالصيغة التربيعيّة

4 u^{2} - 4 u + 1 = 0

الصيغة لحلّ المعادلة التربيعيّة

a = 4, b = - 4, c = 1

لـ

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 4 \cdot 1}{2 \cdot 4}

(- 4)^{2} - 4 \cdot 4 \cdot 1 = 0

(- 4)^{2} - 4 \cdot 4 \cdot 1

زوجيّ n

إذا تحقّق أنّ

, (- a)^{n} = a^{n}

:فعّل قانون القوى

(- 4)^{2} = 4^{2}

= 4^{2} - 4 \cdot 4 \cdot 1

4 \cdot 4 \cdot 1 = 16 :

اضرب الأعداد

= 4^{2} - 16

4^{2} = 16

= 16 - 16

16 - 16 = 0 :

اطرح الأعداد

= 0

u_{1, 2} = \frac{- ( - 4 ) \pm 0}{2 \cdot 4}

u = \frac{- ( - 4 )}{2 \cdot 4}

\frac{- ( - 4 )}{2 \cdot 4} = \frac{1}{2}

\frac{- ( - 4 )}{2 \cdot 4}

- (- a) = a

فعّل القانون

= \frac{4}{2 \cdot 4}

2 \cdot 4 = 8 :

اضرب الأعداد

= \frac{4}{8}

4 :

إلغ العوامل المشتركة

= \frac{1}{2}

u = \frac{1}{2}

حلّ المعادلة التربيعيّة هو

u = \frac{1}{2}

u = cos (x)

استبدل مجددًا

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

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

cos (x) = \frac{1}{2} : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

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

cos (x) = \frac{1}{2} :

حلول عامّة لـ

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

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

وحّد الحلول

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

رسم

أمثلة شائعة

11(1-cos(θ))=sin^2(θ)

11 (1 - cos (θ)) = sin^{2} (θ)

cos^2(x)-1/2 =0,0<= x<= 2pi

cos^{2} (x) - \frac{1}{2} = 0, 0 \leq x \leq 2 π

solvefor x,tan(x)=tan(y)

so l v e f or x, tan (x) = tan (y)

4cos(θ)+3=0

4 cos (θ) + 3 = 0

2cos^2(x)+1=3cos(x),0<= x<= 2pi

2 cos^{2} (x) + 1 = 3 cos (x), 0 \leq x \leq 2 π