solve the equation 4cos^2(x) -15sin(x) = 13

We first want to get every term in terms of the same variable, namely sin(x). to do so, we will use the identity sin^2(x) +cos^2(x) = 1 to get: 4(1-sin^2(x)) -15sin(x) -13 = 0. which we can then rewrite as: 4sin^2(x) +15sin(x)+9=0 and solve it as a quadratic equation in sin(x), giving us: (4sin(x)+3)(sinx+3)=0. Hence x = arcsin(-3/4) or arcsin(-3), Of which only x = arcsin(-3/4) is a valid solution.

MG
Answered by Max G. Further Mathematics tutor

2417 Views

See similar Further Mathematics A Level tutors

Related Further Mathematics A Level answers

All answers ▸

What is the complex conjugate?


y = artanh(x/sqrt(1+x^2)) , find dy/dx


A particle is undergoing circular motion in a horizontal circle, that lies within the smooth surface of a hemispherical bowl of radius 4r. Find the distance OC (explained in diagram) if the angular acceleration of the particle is equal to root (3g/8r).


Using the definitions of hyperbolic functions in terms of exponentials show that sech^2(x) = 1-tanh^2(x)


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning