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# 25sinxcosx-5sinx+20cosx=4

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25sinxcosx-5sinx+20cosx=4

Mar 3, 2015

#3
+94235
+5

Very nice, Anonymous.....

25sinxcosx-5sinx+20cosx= 4

25sinxcosx + 20cosx = 4 + 5sinx

5cosx(5sinx + 4) = (5six + 4)

5cosx = 1

cosx =1/5  at 78.46°

And the other value where sin x = (-4/5) is  about 233°

So...that's all our angles on [0,360]

Mar 3, 2015

#1
+94235
+5

25sinxcosx-5sinx+20cosx= 4

Maybe someone else can solve this one by some algebraic means......but here's a graphical solution.........https://www.desmos.com/calculator/bdgkyrfxh1

It looks like the "x's" that make this true are at 78.5°, 233.1° and 306.9° on the interval [0, 360]

Mar 3, 2015
#2
+5

Partially:

25sinxcosx-5sinx+20cosx=4

5sinx(5cosx-1)=4-20cosx

5sinx= -4(5cosx-1)/ (5cosx-1)

5sinx=-4

sinx=-4/5  which is 306.9 degrees.

I have difficulty with the other values (sleepy).

Mar 3, 2015
#3
+94235
+5

Very nice, Anonymous.....

25sinxcosx-5sinx+20cosx= 4

25sinxcosx + 20cosx = 4 + 5sinx

5cosx(5sinx + 4) = (5six + 4)

5cosx = 1

cosx =1/5  at 78.46°

And the other value where sin x = (-4/5) is  about 233°

So...that's all our angles on [0,360]

CPhill Mar 3, 2015