Question

Sin2x=√2 cos x general solutions? Best expaination will be marked brainliest. For 50 points each

Answer 1

Step-by-step explanation:

$\huge{\underline{\underline{\mathbb{\red{Solution:-}}}}}$

What is the solution of the equation sin(2x)+cos(2x) = 1.2

Expanding on the answer by :

Squaring both sides: (sin(2x)+cos(2x))2=1.22

⇒sin2(2x)+2sin(2x)cos(2x)+cos2(2x)=1.44

As sin2(y)+cos2(y)=1⇒1+2sin(2x)cos(2x)=1.44

Subtracting 1 from both sides of the equation: ⇒2sin(2x)cos(2x)=0.44

As sin(2Y)=sin(Y)cos(Y) , we can rewrite this as: sin(4x)=0.44

Now, two key points to remember:

The sine function returns a positive value in both the first and second quadrants, i.e. sin(y)=a⇒sin(π−y)=a

The sine function is periodic, with a period of 2π , i.e. sin(y)=a⇒sin(y+2kπ)=a , where k is any integer

Using the arcsine function, and expressing angles in radians, we have: 4x=arcsin(0.44)≈0.4556⇒x≈0.1139 is a solution

From 2., we know that 4x+2kπ=arcsin(0.44)≈0.4566

⇒4x≈0.4566−2kπ⇒x≈0.1139−0.5kπ provides one set of solutions.

For example (remembering that k can be negative ):

Using k=−1⇒x≈1.6847

Using k=−2⇒x≈3.2555

Using k=−3⇒x≈4.8263

Don’t forget, we can use any integer value for k .

From 1., we also know that π−4x≈0.4556⇒x≈(0.25π−0.1139)≈0.6715 is also a solution

From 2., we know that π−4x+2kπ=arcsin(0.44)≈0.4566

⇒4x≈(2k+1)π−0.4566⇒x≈(0.5k+0.25)π−0.1139 provides the other set of solutions.

For example:

Using k=1⇒x≈2.2423

Using k=2⇒x≈3.8131

Using k=3⇒x≈5.3839

The two General Solution sets are:

1. x=arcsin(0.44)4+π2k≈0.1139+π2k
2. x=π−arcsin(0.44)4+π2k≈0.6715+pi2k

Answer 2

Answer:

The answer is x= nπ+ (-1) ^n (π/4)