Question

if the sin^2 ( 2x + 45) + cos^2 (3x +20) = 1, tan x =_____

Answer 1

if the $sin^2 ( 2x + 45) + cos^2 (3x +20) = 1$, then x =_____

Answer:

The required answer is $25$.

Step-by-step explanation:

Given: $sin^2 ( 2x + 45) + cos^2 (3x +20) = 1,$

To find: The value of x.

According to question,

$sin^2 ( 2x + 45) + cos^2 (3x +20) = 1,$

Subtract $cos^2 (3x +20)$ on both sides of the issue.

Use the fact that $\sin ^{2} \theta=1-\cos ^{2} \theta$

Use the fact that if $a^{2}=b^{2}$, then $a=\pm b$.

Use the fact that if $\sin x=\sin y$, then $x=n \pi+(-1)^{n} y, n \in \mathbb{Z}$. Hence form an equation in $x$ Solve for $x$

We have

$sin^2 ( 2x + 45) + cos^2 (3x +20) = 1$

Subtracting $cos^2 (3x +20)$ from both sides,

we get $sin^2 ( 2x + 45) = 1- cos^2 (3x +20)$

We know that $\cos ^{2} \theta=1-\sin ^{2} \theta$

Using the above identity, we get $sin^2 ( 2x + 45) = sin^2 (3x +20)$

We know that if $a^{2}=b^{2}$, then $a=\pm b$

Hence, we have $sin( 2x + 45) =\pm sin(3x +20)$

Taking the positive sign, we get $sin( 2x + 45) =sin(3x +20)$

We know that if$\sin x=\sin y$, then $\boldsymbol{x}=n \boldsymbol{\pi }+(-1)^{n} \boldsymbol{y}, n \in \mathcal{B}$,

Hence, we have $( 2x + 45) =n \boldsymbol{\pi }+(-1)^{n}(3x +20)$

Since there are no $\pi$ terms in the options, taking $n=0$,

we get $( 2x + 45) =(3x +20)$ subtracting $45$ on both sides of the issue,

we get $2x=3x-25$

Subtracting $3x$ on both sides of the issue,

we get $x=25$ .

Taking the negative sign, we get the value of $x$ is negative.

Therefore, the value of $x$ is $25$.

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