Solve sin^2pi/4-x=sin^2pi/4+x

$$\sin ^ { 2 } \frac { \pi } { 4 } - x = \sin ^ { 2 } \frac { \pi } { 4 } + x$$

Solve for x

$$x=0$$

Steps for Solving Linear Equation

Get the value of \sin(\frac{\pi }{4}) from trigonometric values table.

$$\left(\frac{\sqrt{2}}{2}\right)^{2}-x=\left(\sin(\frac{\pi }{4})\right)^{2}+x$$

To raise \frac{\sqrt{2}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(\sqrt{2}\right)^{2}}{2^{2}}-x=\left(\sin(\frac{\pi }{4})\right)^{2}+x$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{2^{2}}{2^{2}}.

$$\frac{\left(\sqrt{2}\right)^{2}}{2^{2}}-\frac{x\times 2^{2}}{2^{2}}=\left(\sin(\frac{\pi }{4})\right)^{2}+x$$

Since \frac{\left(\sqrt{2}\right)^{2}}{2^{2}} and \frac{x\times 2^{2}}{2^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(\sqrt{2}\right)^{2}-x\times 2^{2}}{2^{2}}=\left(\sin(\frac{\pi }{4})\right)^{2}+x$$

Get the value of \sin(\frac{\pi }{4}) from trigonometric values table.

$$\frac{\left(\sqrt{2}\right)^{2}-x\times 2^{2}}{2^{2}}=\left(\frac{\sqrt{2}}{2}\right)^{2}+x$$

To raise \frac{\sqrt{2}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(\sqrt{2}\right)^{2}-x\times 2^{2}}{2^{2}}=\frac{\left(\sqrt{2}\right)^{2}}{2^{2}}+x$$

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{2^{2}}{2^{2}}.

$$\frac{\left(\sqrt{2}\right)^{2}-x\times 2^{2}}{2^{2}}=\frac{\left(\sqrt{2}\right)^{2}}{2^{2}}+\frac{x\times 2^{2}}{2^{2}}$$

Since \frac{\left(\sqrt{2}\right)^{2}}{2^{2}} and \frac{x\times 2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(\sqrt{2}\right)^{2}-x\times 2^{2}}{2^{2}}=\frac{\left(\sqrt{2}\right)^{2}+x\times 2^{2}}{2^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{2-x\times 2^{2}}{2^{2}}=\frac{\left(\sqrt{2}\right)^{2}+x\times 2^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{2-x\times 4}{2^{2}}=\frac{\left(\sqrt{2}\right)^{2}+x\times 2^{2}}{2^{2}}$$

Multiply -1 and 4 to get -4.

$$\frac{2-4x}{2^{2}}=\frac{\left(\sqrt{2}\right)^{2}+x\times 2^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{2-4x}{4}=\frac{\left(\sqrt{2}\right)^{2}+x\times 2^{2}}{2^{2}}$$

Divide each term of 2-4x by 4 to get \frac{1}{2}-x.

$$\frac{1}{2}-x=\frac{\left(\sqrt{2}\right)^{2}+x\times 2^{2}}{2^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{1}{2}-x=\frac{2+x\times 2^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{1}{2}-x=\frac{2+x\times 4}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{1}{2}-x=\frac{2+x\times 4}{4}$$

Divide each term of 2+x\times 4 by 4 to get \frac{1}{2}+x.

$$\frac{1}{2}-x=\frac{1}{2}+x$$

Subtract x from both sides.

$$\frac{1}{2}-x-x=\frac{1}{2}$$

Combine -x and -x to get -2x.

$$\frac{1}{2}-2x=\frac{1}{2}$$

Subtract \frac{1}{2} from both sides.

$$-2x=\frac{1}{2}-\frac{1}{2}$$

Subtract \frac{1}{2} from \frac{1}{2} to get 0.

$$-2x=0$$

Product of two numbers is equal to 0 if at least one of them is 0. Since -2 is not equal to 0, x must be equal to 0.

$$x=0$$