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sin^2(x)= 1/9

الحلّ

sin^{2} (x) = \frac{1}{9}

الحلّ

x = 0.33983 \dots + 2 πn, x = π - 0.33983 \dots + 2 πn, x = - 0.33983 \dots + 2 πn, x = π + 0.33983 \dots + 2 πn

+1

درجات

x = 19.47122 \dots^{\circ} + 36 0^{\circ} n, x = 160.52877 \dots^{\circ} + 36 0^{\circ} n, x = - 19.47122 \dots^{\circ} + 36 0^{\circ} n, x = 199.47122 \dots^{\circ} + 36 0^{\circ} n

خطوات الحلّ

sin^{2} (x) = \frac{1}{9}

بالاستعانة بطريقة التعويض

sin^{2} (x) = \frac{1}{9}

sin (x) = u :

على افتراض أنّ

u^{2} = \frac{1}{9}

u^{2} = \frac{1}{9} : u = \frac{1}{3}, u = - \frac{1}{3}

u^{2} = \frac{1}{9}

x = f (a), - f (a)

الحلول هي

x^{2} = f (a)

لـ

u = \frac{1}{9}, u = - \frac{1}{9}

\frac{1}{9} = \frac{1}{3}

\frac{1}{9}

a \geq 0, b \geq 0

بافتراض أنّ

:فعّل قانون الجذور

= \frac{1}{9}

9 = 3

9

9 = 3^{2} :

حلّل العدد لعوامله أوّليّة

= 3^{2}

:فعْل قانون الجذور

3^{2} = 3

= 3

= \frac{1}{3}

1 = 1

فعّل القانون

= \frac{1}{3}

- \frac{1}{9} = - \frac{1}{3}

- \frac{1}{9}

\frac{1}{9}

بسّط:

\frac{1}{3}

\frac{1}{9}

a \geq 0, b \geq 0

بافتراض أنّ

:فعّل قانون الجذور

= \frac{1}{9}

9 = 3

9

9 = 3^{2} :

حلّل العدد لعوامله أوّليّة

= 3^{2}

:فعْل قانون الجذور

3^{2} = 3

= 3

= \frac{1}{3}

= - \frac{1}{3}

1 = 1

فعّل القانون

= - \frac{1}{3}

u = \frac{1}{3}, u = - \frac{1}{3}

u = sin (x)

استبدل مجددًا

sin (x) = \frac{1}{3}, sin (x) = - \frac{1}{3}

sin (x) = \frac{1}{3}, sin (x) = - \frac{1}{3}

sin (x) = \frac{1}{3} : x = arcsin (\frac{1}{3}) + 2 πn, x = π - arcsin (\frac{1}{3}) + 2 πn

sin (x) = \frac{1}{3}

Apply trig inverse properties

sin (x) = \frac{1}{3}

sin (x) = \frac{1}{3} :

حلول عامّة لـ

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

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

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

sin (x) = - \frac{1}{3} : x = arcsin (- \frac{1}{3}) + 2 πn, x = π + arcsin (\frac{1}{3}) + 2 πn

sin (x) = - \frac{1}{3}

Apply trig inverse properties

sin (x) = - \frac{1}{3}

sin (x) = - \frac{1}{3} :

حلول عامّة لـ

sin (x) = - a \Rightarrow x = arcsin (- a) + 2 πn, x = π + arcsin (a) + 2 πn

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

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

وحّد الحلول

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

أظهر الحلّ بالتمثيل العشريّ

x = 0.33983 \dots + 2 πn, x = π - 0.33983 \dots + 2 πn, x = - 0.33983 \dots + 2 πn, x = π + 0.33983 \dots + 2 πn

أمثلة شائعة

2sin^2(x)+sin(x)= 1/2

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

tan (x) = 0.327

(cos(2x))/(sin(x))=1

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

1.04cos^2(θ)+0.102cos(θ)-0.935=0

1.04 cos^{2} (θ) + 0.102 cos (θ) - 0.935 = 0

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

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