Question

cos(3pi/2+x)+sin(3pi/2-x)=0

Answer 1

Answer:

x = pi/ 4 + n*pi  where n is an integer

x = pi /4 - n*pi where n is an integer

Step-by-step explanation:

cos(3pi/2+x)+sin(3pi/2-x)=0

cos (a+b) = cos a cosb - sin a sinb

cos(3pi/2+x) = cos (3pi/2) cos x -sin (3pi/2) sinx

cos 3pi/2 = 0

sin 3pi/2 = -1

cos(3pi/2+x) = cos (3pi/2) cos x -sin (3pi/2) sinx

                    = 0 - (-1) sin x

                     = sin x

sin(3pi/2-x)

sin (a-b) = sina cosb -cosa sin b

sin(3pi/2-x)  = sin (3pi/2) cos x -cos (3pi/2) sin                

cos 3pi/2 = 0

sin 3pi/2 = -1

sin(3pi/2-x)  = sin (3pi/2) cos x -cos (3pi/2) sinx              

                   = -1 *cosx - 0 sinx

                    =-cos x

put these back in

sin x - cos x = 0

add cos x to each side

sin x = cos x

divide by cos x

sin x/cos x = 1

tan x = 1

take the arctan on each side

arctan tan x= arctan 1

x = pi /4

x = pi/ 4 + n*pi  where n is an integer

x = pi /4 - n*pi where n is an integer

Answer 2

$\cos \Bigg(\dfrac{3 \pi}{2} + x \Bigg) + \sin \Bigg(\dfrac{3 \pi}{2} + x \Bigg) = 0$

• Set up

$\cos \Bigg( \dfrac{3 \pi}{2} + x \Bigg) = - \sin \Bigg( \dfrac{3 \pi}{2} + x \Bigg)$

• Subtract $\sin (\frac{3 \pi}{2} + x)$ from both sides

$1 = \dfrac{- \sin \Bigg( \dfrac{3 \pi}{2}  + x \Bigg) }{\cos \Bigg( \dfrac{3 \pi}{2} + x \Bigg )} = - \tan \Bigg( \dfrac{3 \pi}{2} + x \Bigg)$

• Divide both sides by $\cos (\frac{3 \pi}{2} + x)$

$-1 = \tan \Bigg( \dfrac{3 \pi}{2} + x \Bigg)$

• Multiply both sides by -1

For this next step, remember that $\tan ( \frac{3 \pi}{4} ) = -1$ and $\tan (\frac{7 \pi}{4} ) = -1$. You can see that the tangent function is equal to -1 at every $\frac{3 \pi}{4} + n$, where $n$ is an integer. Using substitution, we can say:

$\dfrac{ 3 \pi}{2} + x = \dfrac{3 \pi}{4} + n \pi$

• Rationale explained in the preceeding paragraph

$x = \dfrac{3 \pi}{4} - \dfrac{3 \pi}{2} + n \pi = \dfrac{\pi}{4} + n \pi$

• Subtract $\frac{3 \pi}{2}$ from both sides

Our answer is x = π/4 + nπ, where n is an integer.