Microsoft Math Solver
Solve
Practice
Download
Solve
Practice
Topics
Pre-Algebra
Mean
Mode
Greatest Common Factor
Least Common Multiple
Order of Operations
Fractions
Mixed Fractions
Prime Factorization
Exponents
Radicals
Algebra
Combine Like Terms
Solve for a Variable
Factor
Expand
Evaluate Fractions
Linear Equations
Quadratic Equations
Inequalities
Systems of Equations
Matrices
Trigonometry
Simplify
Evaluate
Graphs
Solve Equations
Calculus
Derivatives
Integrals
Limits
Algebra Calculator
Trigonometry Calculator
Calculus Calculator
Matrix Calculator
Download
Topics
Pre-Algebra
Mean
Mode
Greatest Common Factor
Least Common Multiple
Order of Operations
Fractions
Mixed Fractions
Prime Factorization
Exponents
Radicals
Algebra
Combine Like Terms
Solve for a Variable
Factor
Expand
Evaluate Fractions
Linear Equations
Quadratic Equations
Inequalities
Systems of Equations
Matrices
Trigonometry
Simplify
Evaluate
Graphs
Solve Equations
Calculus
Derivatives
Integrals
Limits
Algebra Calculator
Trigonometry Calculator
Calculus Calculator
Matrix Calculator
Solve
algebra
trigonometry
statistics
calculus
matrices
variables
list
Solve for x
x=\pi n_{1}+\frac{\pi }{4}<br/>n_{1}\in \mathrm{Z}
$x=πn_{1}+4π $
$n_{1}∈Z$
Graph
Graph Both Sides in 2D
Graph in 2D
Quiz
Trigonometry
5 problems similar to:
\sin ( x ) = \cos ( x )
$sin(x)=cos(x)$
Similar Problems from Web Search
How to solve equations like 2 \sin(x) = \cos(x)
How to solve equations like
$2sin(x)=cos(x)$
https://math.stackexchange.com/questions/1476944/how-to-solve-equations-like-2-sinx-cosx/1476973
One way can be using tan\frac x2=t so sin x=\frac{2t}{1+t^2} and cos x=\frac{1-t^2}{1+t^2}. Here 2sin x= cos x implies t^2+4t-1=0 from wich tan \frac x2=2\pm\sqrt{5}.Hence the answer of ...
One way can be using tan
$2x =t$
so sin x=
$1+t_{2}2t $
and cos x=
$1+t_{2}1−t_{2} $
. Here 2sin x= cos x implies
$t_{2}+4t−1=0$
from wich tan
$2x =2±5 $
.Hence the answer of ...
How do you solve \displaystyle{1}+{\sin{{\left({x}\right)}}}={\cos{{\left({x}\right)}}} ?
How do you solve
$1+sin(x)=cos(x)$
?
https://socratic.org/questions/how-do-you-solve-1-sin-x-cos-x
\displaystyle{x}={0} Explanation: \displaystyle{1}+{\sin{{\left({x}\right)}}}={\cos{{\left({x}\right)}}}{\quad\text{or}\quad}{\cos{{x}}}-{\sin{{x}}}={1} . Squaring both sides we get \displaystyle{\left({\cos{{x}}}-{\sin{{x}}}\right)}^{{2}}={1}{\quad\text{or}\quad}{{\cos}^{{2}}{x}}+{{\sin}^{{2}}{x}}-{2}{\sin{{x}}}{\cos{{x}}}={1}{\quad\text{or}\quad}{1}-{\sin{{2}}}{x}={1}{\quad\text{or}\quad}{\sin{{2}}}{x}={0}={\sin{{0}}}; ...
$x=0$
Explanation:
$1+sin(x)=cos(x)orcosx−sinx=1$
. Squaring both sides we get
$(cosx−sinx)_{2}=1orcos_{2}x+sin_{2}x−2sinxcosx=1or1−sin2x=1orsin2x=0=sin0;$
...
How do you show that the equation \displaystyle{1}-{\sin{{x}}}={\cos{{x}}} is not an identity?
How do you show that the equation
$1−sinx=cosx$
is not an identity?
https://socratic.org/questions/how-do-you-show-that-the-equation-1-sinx-cosx-is-not-an-identity
Bdub Nov 12, 2016 Pick a value for x like \displaystyle\frac{\pi}{{3}} and plug it in to both side to show that they don't equal each other and therefore not an identity
Bdub Nov 12, 2016 Pick a value for x like
$3π $
and plug it in to both side to show that they don't equal each other and therefore not an identity
Trigonometric equation \sin2x=\cos x
Trigonometric equation
$sin2x=cosx$
https://math.stackexchange.com/questions/3008492/trigonometric-equation-sin2x-cos-x
As @Nicholas Stull hinted, you lost solutions by not making sure that you were not dividing by zero. As @Winther pointed out, you can avoid this error by factoring. As @Nicholas Stull pointed out, ...
As @Nicholas Stull hinted, you lost solutions by not making sure that you were not dividing by zero. As @Winther pointed out, you can avoid this error by factoring. As @Nicholas Stull pointed out, ...
If \sin(x)=3\cos(x), compute \sin(x)*\cos(x) [closed]
If
$sin(x)=3cos(x)$
, compute
$sin(x)∗cos(x)$
[closed]
https://math.stackexchange.com/questions/2685713/if-sinx-3-cosx-compute-sinx-cosx
We have \tan(x) = 3 Draw a right angle triangle where x is one of the angle, let the opposite side be of length 3, let the adjacent side be of length one. Then by Pythagoras Theorem, the ...
We have
$tan(x)=3$
Draw a right angle triangle where
$x$
is one of the angle, let the opposite side be of length
$3$
, let the adjacent side be of length one. Then by Pythagoras Theorem, the ...
Find \sin x \cos x if \sin x = 4 \cos x
Find
$sinxcosx$
if
$sinx=4cosx$
https://math.stackexchange.com/questions/1092322/find-sin-x-cos-x-if-sin-x-4-cos-x
Hint Maybe starting with \sin^2 x + \cos^2 x = 1 is helpful :)
Hint Maybe starting with
$sin_{2}x+cos_{2}x=1$
is helpful :)
More Items
Share
Copy
Copied to clipboard
Similar Problems
\tan ( x )
$tan(x)$
\sec ( x )
$sec(x)$
\sin ( x ) = \cos ( x )
$sin(x)=cos(x)$
\cot ( x )
$cot(x)$
\cos ( x )
$cos(x)$
\csc ( x )
$csc(x)$
Back to top