SOLUTION: Solve 12sin^2(w)+7cos(w)−13=0 for all solutions 0 ≤w<2π

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Question 1157544: Solve 12sin^2(w)+7cos(w)−13=0 for all solutions 0 ≤w<2π
Found 2 solutions by Edwin McCravy, josmiceli:
Answer by Edwin McCravy(20065) About Me  (Show Source):
You can put this solution on YOUR website!
12sin%5E2%28w%29%2B7cos%28w%29-13+=+0

12%281-cos%5E2%28w%29%5E%22%22%29%2B7cos%5E%22%22%28w%29+-+13+=+0

12-12cos%5E2%28w%29%2B7cos%5E%22%22%28w%29+-+13+=+0

-12cos%5E2%28w%29%2B7cos%5E%22%22%28w%29+-+1+=+0

12cos%5E2%28w%29-7cos%5E%22%22%28w%29+%2B+1+=+0

%284cos%5E%22%22%28w%29-1%29%283cos%5E%22%22%28w%29-1%29=0

4cos(w)-1=0;  3cos(w)-1=0
  4cos(w)=1;    3cos(w)=1
   cos(w)=1/4;   cos(w)=1/3

Those are positive numbers.  The cosine is positive in 
quadrants QI and QIV, so

For cos(w)=1/4,
w = 1.318116072 radians in QI, 2π-1.318116072 = 4.965069235 in QIV

For cos(w)=1/3,
w = 1.230959417 radians in QI, 2π-1.230959417 = 5.05222589 in QIV

Edwin

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+12%2Asin%5E2%28w%29+%2B+7%2Acos%28w%29+-+13+=+0+
+sin%5E2%28w%29+=+1+-+cos%5E2%28w%29+
+12%2A%28+1+-+cos%5E2%28w%29+%29+%2B+7%2Acos%28w%29+-+13+=+0+
+12+-+12%2Acos%5E2%28w%29+%2B+7%2Acos%28w%29+-+13+=+0+
+12%2Acos%5E2%28w%29+-+7%2Acos%28w%29+%2B+1+=+0+
Let +z+=+cos%28w%29+
+12z%5E2+-+7z+%2B+1+=+0+
+%28+4z+-+1+%29%2A%28+3z+-+1+%29+=+0+ ( by looking at it )
+4z+=+1+
+z+=+1%2F4+
and
+3z+=+1+
+z+=+1%2F3+
-----------------
+cos%28w%29+=+1%2F4+
The cosine is positive in 1st and 4th quadrants
+w+=+arc+cos%28+1%2F4+%29+
+w+=+1.3181+ radians
+w+=+2pi+-+1.3181+=+4.9651+ radians
and
+cos%28w%29+=+1%2F3+
+w+=+arc+cos%28+1%2F3+%29+
+w+=+1.2310+ radians
+w+=+2pi+-+1.2310+=+5.0522+ radians
----------------------------------------------
check answers (4)
+w+=+1.3181+
+12%2Asin%5E2%28w%29+%2B+7%2Acos%28w%29+-+13+=+0+
+12%2A.93749+%2B+7%2A.25+-+13+=+0+
+11.25+%2B+1.75+-+13+=+0+
+13+-+13+=+0+
OK
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