Question

Solve for x: 2sinx + 2cosx = 1

Answer 1

for \(a\sin x+b\cos x=c\),
divide both sides by \(\sqrt{a^2+b^2}\)

Answer 2

then use the formula, for sin Acos B+cos A sin B = sin (A+B)

Answer 3

did u try this ?

Answer 4

how can I apply the formula sin Acos B+cos A sin B = sin (A+B)?

Answer 5

then \(\cos \alpha =a/\sqrt{a^2+b^2}\)
and \(\sin\alpha =b/\sqrt{a^2+b^2}\)
so u get \(\cos \alpha \sin x+\sin \alpha \cos x= c/\sqrt{a^2+b^2} \\\sin(x+\alpha )=c/\sqrt{a^2+b^2}\)

Answer 6

in right side, that is 1 or 2 ?

Answer 7

1

Answer 8

I'm sorry, I'm lost right here...

Answer 9

\(x+\alpha =\sin^{-1}(c/\sqrt{a^2+b^2})\\ x=-\alpha +\sin^{-1}(c/\sqrt{a^2+b^2})\)

Answer 10

i've solved for 'x' for any general expression, \(a\sin x+b\cos x=c\)
tell me which step you have doubt with ?

Answer 11

But what does α equal to?

\(\cos^{-1}{\left(\dfrac{a}{\sqrt{a^2 + b^2}}\right)}\) ?

Answer 12

yes, thats correct.

Answer 13

Wow, it looks ugly, but thanks.

Answer 14

or arctan(b/a)

Answer 15

ah, much better.

Answer 16

if my way :
2sinx + 2cosx = 1
divide both sides by 2
sinx + cosx = 1/2
square both sides
sin^2 x + cos^2 x + 2sinxcosx = 1/4
1 + sin2x = 1/4
sin2x = 1/4 - 1
sin2x = -3/4
actually, to get x in right side give a familiar value in sine of special angles
that's why i ask before, in right side 1 or 2

Answer 17

but how did you get arctan(b/a) ?

Answer 18

never mind, i got it.

Answer 19

if we solve using numbers, it isn't that ugly....
sin alpha/ tan alpha =.... ?

Answer 20

squaring both sides may add extraneous roots...