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Question
HOw do I solve an equation like this
sin X = - cos x
sinx=-cosx --> snix+cosx=0 . Let's set sinx=t . We know that sin²x+cos²=1 therefore :
cosx=√(1-sin²x) , thus cosx=√(1-t²). Hence, out equation becomes :
t+√(1-t²)=0
t=-√(1-t²)
t²=1-t²
2t²=1
t=±1/√2 .
# t=1/√2 , that means sinx=1/√2 -- > x=π/4 & x=3π/4 +2πk , k=0,1,2,3,4,...
# t=-1/√2 , that means sinx=-1/√2 -- > x=5π/4 & x=7π/4 +2πk , k=0,1,2,3,4,...
Therefore : sinx+cosx=0 when x=π(1+2k)/4 , k=0,1,2,3,4,5,6,....
Alon.
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Answers by Expert:
Alon Mandes
Expertise
Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems. Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .
Experience
1. I'm a team member of mathnerds (math site for answering questions)
2. I'm a team member in the Student's Union of the Technion, helping
students who have problems in mathematics.
3. 2 years of experience as a math teacher in college.
4. I give free homework help for high school students in
Mathematics & Physics.
5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" ,
"Complex Functions".
Organizations
Hi-Tech company : GSM4VOIP ; job possition : Algorythm developer.
Education/Credentials
M.A in Mathematics & Bs.c in Electronics.
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