What is the general solution for solvefor c,cos(4x-90)= 1/s ecx ?

The general solution for solvefor c,cos(4x-90)= 1/s ecx is c=(cos(4x-90)s)/(ex);\quad s\ne 0,x\ne 0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

solvefor c,cos(4x-90)= 1/s ecx

solve for

c, cos (4 x - 90) = \frac{1}{s} ec x

c = \frac{cos ( 4 x - 90 ) s}{e x}; s \neq = 0, x \neq = 0

Solution steps

cos (4 x - 90) = \frac{1}{s} ec x

Switch sides

\frac{1}{s} ec x = cos (4 x - 90)

Multiply both sides by

s

\frac{1}{s} ec x = cos (4 x - 90)

Multiply both sides by

s

\frac{1}{s} ec x s = cos (4 x - 90) s; s \neq = 0

Simplify

ec x = cos (4 x - 90) s; s \neq = 0

ec x = cos (4 x - 90) s; s \neq = 0

Divide both sides by

e x; s \neq = 0, x \neq = 0

ec x = cos (4 x - 90) s; s \neq = 0

Divide both sides by

e x; s \neq = 0, x \neq = 0

\frac{ec x}{e x} = \frac{cos ( 4 x - 90 ) s}{e x}; s \neq = 0, x \neq = 0

Simplify

c = \frac{cos ( 4 x - 90 ) s}{e x}; s \neq = 0, x \neq = 0

c = \frac{cos ( 4 x - 90 ) s}{e x}; s \neq = 0, x \neq = 0

i, 2 (1 + sin (x)) cos^{2} (2 x) = cos^{2} (x)

i, cos (x + 4 5^{\circ}) = 5 sin (x)

c, cos^{2} (x) + cos (x) = 1 + sin^{2} (x) - sin (x)

\frac{cos ^{2} ( x ) - 1}{4} = 0

cos (x) + cos (3 x) = \frac{1}{2}

What is the general solution for solvefor c,cos(4x-90)= 1/s ecx ?
The general solution for solvefor c,cos(4x-90)= 1/s ecx is c=(cos(4x-90)s)/(ex);\quad s\ne 0,x\ne 0