Question

Help now for brainliest,5stars and thanks.Find the values of x in the range 0≤x≤180° for which 15sin^2 x+11cox-17=0

Answer 1

Answer:

$\bold{x = 66.42^\circ, 70.53^\circ}$

Step-by-step explanation:

Given equation is:

$15sin^2 x+11cosx-17=0$

To find:

Values of $x$ in the range $0^\circ\leq x\leq180^\circ$.

Solution:

First of all, let us recall one identity of square of sine and square of cosine.

Sum of square of sine of an angle and square of cosine of the same angle is equal to 1.

$\bold{sin^2\theta+cos^2\theta=1}$

So

Putting $sin^2x=1-cos^2x$ in the given equation.

$15(1-cos^2x)+11cosx-17=0\\\Rightarrow 15-15cos^2x+11cosx-17=0\\\Rightarrow -15cos^2x+11cosx-2=0\\\Rightarrow 15cos^2x-11cosx+2=0\\\Rightarrow 15cos^2x-5cosx-6cosx+2=0\\\Rightarrow 5cosx(3cosx-1)-2(3cosx-1)=0\\\Rightarrow (5cosx-2)(3cosx-1)=0\\\Rightarrow cosx=\dfrac{2}{5}, \dfrac{1}{3}$

So, in the range $0^\circ\leq x\leq180^\circ$

$\bold{x = 66.42^\circ, 70.53^\circ}$