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شائع علم المثلثات >

sec^2(x)-2=tan^2(x)

الحلّ

sec^{2} (x) - 2 = tan^{2} (x)

الحلّ

x \in R لا يوجد حلّ لـ

خطوات الحلّ

sec^{2} (x) - 2 = tan^{2} (x)

من الطرفين

tan^{2} (x)

اطرح

sec^{2} (x) - 2 - tan^{2} (x) = 0

sin, cos :

عبّر بواسطة

- 2 + sec^{2} (x) - tan^{2} (x)

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

:Use the basic trigonometric identity

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

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

:Use the basic trigonometric identity

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

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

بسّط:

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

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

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

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

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

:فعّل قانون القوى

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

1^{a} = 1

فعّل القانون

1^{2} = 1

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

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

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

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

:فعّل قانون القوى

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

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

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

وحّد الكسور:

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

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

فعّل القانون

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

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

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

:حوّل الأعداد لكسور

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

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

:بما أنّ المقامات متساوية، اجمع البسوط

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

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

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

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

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

Rewrite using trig identities

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

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

:فعّل نطريّة فيتاغوروس

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

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

بسّط

= - cos^{2} (x)

- cos^{2} (x) = 0

- 1

اقسم الطرفين على

- cos^{2} (x) = 0

- 1

اقسم الطرفين على

- cos^{2} (x) = 0

- 1

اقسم الطرفين على

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

بسّط

cos^{2} (x) = 0

cos^{2} (x) = 0

x^{n} = 0 \Rightarrow x = 0

فعّل القانون

cos (x) = 0

cos (x) = 0 :

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}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn

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

\frac{π}{2} + 2 πn, \frac{3 π}{2} + 2 πn :

بما أنّ المعادلة غير معرّفة لـ

x \in R لا يوجد حلّ لـ

رسم

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tan(x)=0.5

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sec^2(x)-3sec(x)=0

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