What is the general solution for sin(x+pi/4)=sqrt(2)cos(x+pi/4) ?

The general solution for sin(x+pi/4)=sqrt(2)cos(x+pi/4) is x=(0.33983…)/2+pin

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sin(x+pi/4)=sqrt(2)cos(x+pi/4)

sin (x + \frac{π}{4}) = 2 cos (x + \frac{π}{4})

x = \frac{0.33983 \dots}{2} + πn

Solution steps

sin (x + \frac{π}{4}) = 2 cos (x + \frac{π}{4})

Square both sides

sin^{2} (x + \frac{π}{4}) = (2 cos (x + \frac{π}{4}))^{2}

Rewrite using trig identities

\frac{( sin ( x ) + cos ( x ) ) ^{2}}{2} = (cos (x) - sin (x))^{2}

Subtract

(cos (x) - sin (x))^{2}

from both sides

\frac{- sin ^{2} ( x ) + 6 cos ( x ) sin ( x ) - cos ^{2} ( x )}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- sin^{2} (x) + 6 cos (x) sin (x) - cos^{2} (x) = 0

Rewrite using trig identities

- 1 + 6 \cdot \frac{sin ( 2 x )}{2} = 0

6 \cdot \frac{sin ( 2 x )}{2} = 3 sin (2 x)

- 1 + 3 sin (2 x) = 0

Move

1

to the right side

3 sin (2 x) = 1

Divide both sides by

3

sin (2 x) = \frac{1}{3}

Apply trig inverse properties

2 x = arcsin (\frac{1}{3}) + 2 πn, 2 x = π - arcsin (\frac{1}{3}) + 2 πn

Solve

2 x = arcsin (\frac{1}{3}) + 2 πn : x = \frac{arcsin ( \frac{1}{3} )}{2} + πn

Solve

2 x = π - arcsin (\frac{1}{3}) + 2 πn : x = \frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + πn

x = \frac{arcsin ( \frac{1}{3} )}{2} + πn, x = \frac{π}{2} - \frac{arcsin ( \frac{1}{3} )}{2} + πn

Verify solutions by plugging them into the original equation

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

Show solutions in decimal form

x = \frac{0.33983 \dots}{2} + πn

5 = 7 cos (\frac{π}{3} t)

6 cos^{2} (x) - 4 = 0

csc (x) = 3.5

sin (x) = 0, sin (x) = 0

cos (x) = - \frac{7}{9}, \frac{π}{2} < x < π

What is the general solution for sin(x+pi/4)=sqrt(2)cos(x+pi/4) ?
The general solution for sin(x+pi/4)=sqrt(2)cos(x+pi/4) is x=(0.33983…)/2+pin