Solve for x: sin^3 x - cos^3 x = 1 + sin x * cos x
#solve this trigonometric equation
timber citrus
foggy bronze
Solve for x: sin^3 x - cos^3 x = 1 + sin x * cos x
for a and b real numbers factor out a^3-b^3 (ie use Bernoulli factorisation ) then use some trig identities
timber citrus
can you solve it please
foggy bronze
can you solve it please
No, I can help you to solve it but I won’t solve it for you is this for school ?
timber citrus
No, I can help you to solve it but I won’t solve it for you is this for school ?
yes
foggy bronze
Well I don’t think this is an appropriate channel then
If you need help you should post it here
And helpers are here to help, not solve, solving questions without you understanding them is not helping
timber citrus
I have solved it but answer is not matching
foggy bronze
Okay, what did you do ?
timber citrus
apply formula
foggy bronze
Which formula ?
foggy bronze
a³-b³=(a-b)(a²+b²+ab)
Yea indeed that’s what you need to do and here a^2+b^2=1 using trig
timber citrus
what after that
foggy bronze
So you have (cos(x)-sin(x))(1+cos(x)sin(x))=1+cos(x)sin(x)
timber citrus
ya
foggy bronze
Well then factor by 1+cos(x)sin(x) and from there you will have two possibilities
timber citrus
then sinx-cosx=1
or 1+sinx= 0 ?
foggy bronze
Well factoring on each side either cos(x)-sin(x)=1 or 1+cos(x)sin(x)=0
The second equation has no solutions so you only have to treat the first one
Hmm wait it’s actually sin(x)-cos(x)=1 mb
timber citrus
Hmm wait it’s actually sin(x)-cos(x)=1 mb
ya...
twin zinc
Hey, you got it?
timber citrus
dayumn
dapper sphinx
wind orchid
yeah where is my ex
💀
mortal shore
ya...
I got x = π, π/2 for sinx - cosx = 1. π yields sin(π) - cos(π) —> 0 - (-1) = 1. π/2 yields sin(π/2) - cos(π/2) —> 1 - 0 = 1.
I didn’t draw a triangle or use inverses, I just checked every 45 degrees on the unit circle for which worked
I checked and the full answer is x = π + 2kπ k∈ℤ, π/2 + 2kπ k∈ℤ