الحل العام للمعادلة المثلثية اهداء الاستاذ ابراهيم الاحمدى ابراهيم

1. 
‫ا‬‫ﻟﺤﻞ‬‫اﻟﻌﺎم‬
‫ﻟ‬‫ﻠﻤﻌﺎدﻟﺔ‬‫اﻟﻤﺜﻠﺜﻴﺔ‬
‫ﺗ‬‫ــ‬‫ﺄﻟﻴ‬‫ـ‬‫ﻒ‬



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2. etn.alafdal.4mathtop.www1‫اﻟﻤﺘﻘﺪﻣﺔ‬ ‫اﻟﺮﯾﺎﺿﯿﺎت‬ ‫ﻣﻨﺘﺪى‬
0112 093 0112 ‫اﻟﺮﯾﺎﺿﯿﺎت‬ ‫ﺧﺒﯿﺮ‬ ‫ﻣﻌﻠﻢ‬ ‫اﺑﺮاھﯿﻢ‬ ‫اﻷﺣﻤﺪى‬ ‫اﺑﺮاھﯿﻢ‬ / ‫اﻷﺳﺘﺎذ‬ ‫اﻋﺪاد‬
‫اﻟﻤﺜﻠﺜﯿﺔ‬ ‫ﻟﻠﻤﻌﺎدﻟﺔ‬ ‫اﻟﻌﺎم‬ ‫اﻟﺤﻞ‬
β‫ﺟﺎ‬ = α ‫ﺟﺎ‬ ‫اﻟﺼﻮرة‬ ‫ﻋﻠﻰ‬ ‫اﻟﺘﻰ‬ ‫اوﻻ:اﻟﻤﻌﺎدﻟﺔ‬
‫ﯾﻜﻮن‬ ‫اﻟﻌﺎم‬ ‫اﻟﺤﻞ‬α=‫ﻥﺐ‬+)-1(‫ﻥ‬β
‫اﻟﺒﺮھﺎن‬
‫اﻣﺎ‬‫ﺟﺎ‬α=‫ﺟﺎ‬β‫ﺉ‬α=٢‫ﻥﺐ‬+β
‫ﺉ‬α=‫زوﺟﻰ‬ ‫ﻋﺪد‬‫×ﺐ‬+β
‫ا‬‫و‬‫ﺟﺎ‬α=‫ﺟﺎ‬)‫ﺐ‬-β(‫ﺉ‬α=٢‫ﻥﺐ‬+‫ﺐ‬-β
‫ﺉ‬α) =٢‫ﻥ‬+1(‫ﺐ‬-β
‫ﺉ‬α=‫ﻓﺮدى‬ ‫ﻋﺪد‬‫×ﺐ‬-β
‫ﺇ‬‫اﻟﺤﻞ‬‫ﯾﻜﻮن‬ ‫اﻟﻌﺎم‬α=‫ﻥﺐ‬+)-1(‫ﻥ‬β
‫ﺗﻄﺒﯿﻘﻰ‬ ‫ﻣﺜﺎل‬
‫ﻟﻠﻤﻌﺎدﻟﺔ‬ ‫اﻟﻌﺎم‬ ‫اﻟﺤﻞ‬ ‫أوﺟﺪ‬:‫ﺟﺎ‬٢θ=‫ﺟ‬‫ﺘ‬‫ﺎ‬θ
‫اﻟﺤﻞ‬
‫ﺟﺎ‬٢θ=‫ﺟﺎ‬)-θ(
٢θ=‫ﻥﺐ‬+)-1(‫ﻥ‬)-θ(
‫ﻛﺎﻧﺖ‬ ‫اذا‬‫ن‬‫ﻓﺎن‬ ‫زوﺟﻰ‬٢θ=‫ﻥﺐ‬) +-θ(
٣θ=‫ﻥﺐ‬+‫ﺉ‬θ+ =
‫ﻛﺎﻧﺖ‬ ‫اذا‬‫ن‬‫ﻓﺮدى‬‫ﻓﺎن‬٢θ=‫ﻥﺐ‬-)-θ(
٢θ=‫ﻥﺐ‬-+θ‫ﺉ‬θ=‫ﻥﺐ‬-
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3. etn.alafdal.4mathtop.www2‫اﻟﻤﺘﻘﺪﻣﺔ‬ ‫اﻟﺮﯾﺎﺿﯿﺎت‬ ‫ﻣﻨﺘﺪى‬
0112 093 0112 ‫اﻟﺮﯾﺎﺿﯿﺎت‬ ‫ﺧﺒﯿﺮ‬ ‫ﻣﻌﻠﻢ‬ ‫اﺑﺮاھﯿﻢ‬ ‫اﻷﺣﻤﺪى‬ ‫اﺑﺮاھﯿﻢ‬ / ‫اﻷﺳﺘﺎذ‬ ‫اﻋﺪاد‬
β‫ﺟﺘﺎ‬ = α ‫ﺟﺘﺎ‬ ‫اﻟﺼﻮرة‬ ‫ﻋﻠﻰ‬ ‫اﻟﺘﻰ‬ ‫ﺛﺎﻧﯿﺎ:اﻟﻤﻌﺎدﻟﺔ‬
‫ﯾﻜﻮن‬ ‫اﻟﻌﺎم‬ ‫اﻟﺤﻞ‬α=٢‫ﻥﺐ‬±β
‫اﻟﺒﺮھﺎن‬
‫اﻣﺎ‬‫ﺟ‬‫ﺘ‬‫ﺎ‬α=‫ﺟ‬‫ﺘ‬‫ﺎ‬β‫ﺉ‬α=٢‫ﻥﺐ‬+β
‫ﺉ‬α=‫ﻋﺪد‬‫زوﺟﻰ‬‫×ﺐ‬+β
‫او‬‫ﺟ‬‫ﺘ‬‫ﺎ‬α=‫ﺟ‬‫ﺘ‬‫ﺎ‬)٢‫ﺐ‬-β(‫ﺉ‬α=٢‫ﻥﺐ‬+٢‫ﺐ‬-β
‫ﺉ‬α) =٢‫ﻥ‬+٢(‫ﺐ‬-β
‫ﺉ‬α=‫ﻋﺪد‬‫زوﺟﻰ‬‫×ﺐ‬-β
‫ﺇ‬‫اﻟﺤﻞ‬‫ﯾﻜﻮن‬ ‫اﻟﻌﺎم‬α=٢‫ﻥﺐ‬±β
‫ﺗﻄﺒﯿﻘﻰ‬ ‫ﻣﺜﺎل‬
‫ﻟﻠﻤﻌﺎدﻟﺔ‬ ‫اﻟﻌﺎم‬ ‫اﻟﺤﻞ‬ ‫أوﺟﺪ‬:‫ﺟﺎ‬٢θ=‫ﺟﺘﺎ‬θ
‫اﻟﺤﻞ‬
‫ﺟ‬‫ﺘ‬‫ﺎ‬)-٢θ(=‫ﺟ‬‫ﺘ‬‫ﺎ‬θ
θ=٢‫ﻥﺐ‬±)-٢θ(
‫اﻣﺎ‬θ=٢‫ﻥﺐ‬) +-٢θ(
٣θ=٢‫ﻥﺐ‬+‫ﺉ‬θ=+
‫او‬θ=٢‫ﻥﺐ‬-)-٢θ(
θ=٢‫ﻥﺐ‬-+٢θ
-θ=٢‫ﻥﺐ‬-‫ﺉ‬θ=-٢‫ﻥﺐ‬+
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4. etn.alafdal.4mathtop.www3‫اﻟﻤﺘﻘﺪﻣﺔ‬ ‫اﻟﺮﯾﺎﺿﯿﺎت‬ ‫ﻣﻨﺘﺪى‬
0112 093 0112 ‫اﻟﺮﯾﺎﺿﯿﺎت‬ ‫ﺧﺒﯿﺮ‬ ‫ﻣﻌﻠﻢ‬ ‫اﺑﺮاھﯿﻢ‬ ‫اﻷﺣﻤﺪى‬ ‫اﺑﺮاھﯿﻢ‬ / ‫اﻷﺳﺘﺎذ‬ ‫اﻋﺪاد‬
β‫ظﺎ‬ = α ‫ظﺎ‬ ‫اﻟﺼﻮرة‬ ‫ﻋﻠﻰ‬ ‫اﻟﺘﻰ‬ ‫ﺛﺎﻟﺜﺎ:اﻟﻤﻌﺎدﻟﺔ‬
‫ﯾﻜﻮن‬ ‫اﻟﻌﺎم‬ ‫اﻟﺤﻞ‬α=‫ﻥﺐ‬+β
‫اﻟﺒﺮھﺎن‬
‫اﻣﺎ‬‫ظ‬‫ﺎ‬α=‫ظ‬‫ﺎ‬β‫ﺉ‬α=٢‫ﻥﺐ‬+β
‫ﺉ‬α=‫ﻋﺪد‬‫زوﺟﻰ‬‫×ﺐ‬+β
‫او‬‫ظ‬‫ﺎ‬α=‫ظ‬‫ﺎ‬)‫ﺐ‬+β(‫ﺉ‬α=٢‫ﻥﺐ‬+‫ﺐ‬+β
‫ﺉ‬α) =٢‫ﻥ‬+1(‫ﺐ‬+β
‫ﺉ‬α=‫ﻋﺪد‬‫ﻓﺮدى‬+β
‫ﺇ‬‫اﻟﺤﻞ‬‫ﯾﻜﻮن‬ ‫اﻟﻌﺎم‬α=‫ﻥﺐ‬+β
‫ﺗﻄﺒﯿﻘﻰ‬ ‫ﻣﺜﺎل‬
‫ﻟﻠﻤﻌﺎدﻟﺔ‬ ‫اﻟﻌﺎم‬ ‫اﻟﺤﻞ‬ ‫أوﺟﺪ‬:‫ظﺎ‬٢θ=‫ظﺘﺎ‬θ
‫اﻟﺤﻞ‬
‫ظﺎ‬٢θ=‫ظﺎ‬)-θ(
٢θ=‫ﻥﺐ‬) +-θ(
٣θ=‫ﻥﺐ‬+
θ+ =
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٢‫ﺐ‬
٢‫ﺐ‬
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